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-{"url": "https://math.stackexchange.com/questions/4082284/solving-for-vector-contained-in-a-diagonal-matrix", "content": "Consider the following system of equations:\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{z}D_{\\mathbf{y}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere $\\mathbf{x}$ , $\\mathbf{y}$ , $\\mathbf{z}$ , and $\\mathbf{u}$ are $1\\times n$ vectors and $D_{\\mathbf{x}}$ , $D_{\\mathbf{y}}$ , and $D_{\\mathbf{z}}$ are diagonal $n\\times n$ matrices with $\\mathbf{x}$ , $\\mathbf{y}$ , and $\\mathbf{z}$ , respectively, along their diagonals (i.e., $D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})$ ).\n\nMy question is whether it is possible to solve for $\\mathbf{y}$ here – both the $\\mathbf{y}$ as a vector and the $\\mathbf{y}$ along the diagonal of $D_{\\mathbf{y}}$ . The problem is that I do not know of operations to pull $\\mathbf{y}$ out of $D_{\\mathbf{y}}$ . The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write\n\n$$\n(\\mathbf{y}e^T) \\odot I_n = D_{\\mathbf{y}}\n$$\n\nwhere $\\odot$ is the Hadamard product and $e^T = (1,1,\\ldots)\\in\\mathbb R^n$ . So, would it be possible to distribute out $\\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\\mathbf{y}$ , not $D_{\\mathbf{y}}$ :\n\n$$\n\\mathbf{y}(e^T \\odot I_n) \\neq D_{\\mathbf{y}}\n$$\n\nEdit: Oh, it seems that extracting $\\mathbf{y}$ in this case would be a simple as rewriting the equation above as\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{y}D_{\\mathbf{z}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nbecause rewriting the equation this way would not change the terms along the diagonal of $\\mathbf{z}D_{\\mathbf{y}}$ . Then we can write\n\n$$\n\\mathbf{y}=\\mathbf{u}D_{\\mathbf{z}}D_{\\mathbf{x+z}}^{-1}\n$$\n\nBut then how would solve for $\\mathbf{y}$ in the following?\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{y}D_{\\mathbf{y}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nI think that in this case, it would not be possible to solve for a single vector $\\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:\n\n$$\n\\mathbf{y^2} + \\mathbf{y}D_{\\mathbf{x}} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere $\\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\\mathbf{y}$ – that is,\\$\\mathbf{y^2} =\n\n$$\n\\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}\n$$\n\n\\$.\n\nNext, what if we had an equation as follows?\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{y}D_{\\mathbf{yM}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere $\\mathbf{M}$ is an $n \\times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\\mathbf{yM}}$ , we have some linear combination.\n\nI think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\\mathbf{M}$ and $D_{\\mathbf{x}}$ that is multiplied by a scalar ( $n$ , I believe). Then, we would have a set of vectors ( $n-1$ , I believe) that each have as elements different products of the elements in $\\mathbf{y}$ (e.g., $y_1 y_3$ ) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\\mathbf{M}$ and $D_{\\mathbf{x}}$ .\n\nSo, I have two questions here:\n\n1. Is my general intuition about what this equation would look like correct?\n2. Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?\n\n## 1 Answer\n\n1\n\nI am not sure what the notation $D_\\mathbf{x}$ , $D_\\mathbf{y}$ , and $D_\\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\\mathbf{x}}$ is invertible, you may solve for $\\mathbf{y}$ as follows given the original equation:\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{z}D_{\\mathbf{y}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}=\\mathbf{u}D_{\\mathbf{z}}-\\mathbf{z}D_{\\mathbf{y}}\n$$\n\n$$\n\\mathbf{y}=\\bigr( \\mathbf{u}D_{\\mathbf{z}}-\\mathbf{z}D_{\\mathbf{y}} \\bigr) D_{\\mathbf{x}}^{-1}\n$$\n\nNote: I can't verify that your edit is correct because I don't understand what $D_{\\mathbf{x} + \\mathbf{z}}$ represents.\n\nEdit 1:\n\nThank you for the response as I didn't realize the meaning of the matrices $D_{\\mathbf{x}}$ , $D_{\\mathbf{y}}$ , and $D_{\\mathbf{z}}$ . Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.\n\nSince $D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})$ and similarly for the other matrices, we have\n\n$$\n\\begin{bmatrix} y_1 & \\cdots & y_n \\end{bmatrix} \\begin{bmatrix} x_1 & & \\\\ & \\ddots & \\\\ & & x_n\\end{bmatrix} + \\begin{bmatrix} z_1 & \\cdots & z_n \\end{bmatrix} \\begin{bmatrix} y_1 & & \\\\ & \\ddots & \\\\ & & y_n\\end{bmatrix} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nand multiplying through we have\n\n$$\n\\begin{align}\n\\begin{bmatrix} y_1 x_1 & \\cdots & y_n x_n \\end{bmatrix} + \\begin{bmatrix} y_1 z_1 & \\cdots & y_n z_n \\end{bmatrix} &= \\mathbf{u}D_{\\mathbf{z}}\\\\\n\\begin{bmatrix} y_1 x_1 + y_1 z_1 & \\cdots & y_n x_n + y_n z_n \\end{bmatrix} &= \\mathbf{u}D_{\\mathbf{z}} \\\\\n\\begin{bmatrix} y_1 (x_1 + z_1) & \\cdots & y_n(x_n + z_n) \\end{bmatrix} &= \\mathbf{u}D_{\\mathbf{z}} \\\\\n\\end{align}\n$$\n\nTherefore, this can be written in matrix form as\n\n$$\n\\begin{bmatrix} y_1 & \\cdots & y_n \\end{bmatrix} \\begin{bmatrix} x_1 + z_1 & & \\\\ & \\ddots & \\\\ & & x_n + z_n \\end{bmatrix} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nor more concisely as\n\n$$\n\\mathbf{y} (D_{\\mathbf{x}} + D_{\\mathbf{z}} ) = \\mathbf{u} D_{\\mathbf{z}}\n$$\n\nwhich is exactly what is given in your edit:\n\n$$\n\\mathbf{y} = \\mathbf{u} D_{\\mathbf{z}} (D_{\\mathbf{x}} + D_{\\mathbf{z}} ) ^{-1}\n$$\n\nEdit 2:\n\nAs for the case where you have $\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}$ where\\$\\mathbf{y}^2 =\n\n$$\n\\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}\n$$\n\n\\$, you would not be able to solve for $\\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):\n\n$$\n\\begin{bmatrix} x_1 & \\cdots & x_n \\end{bmatrix} \\begin{bmatrix} y_1^2 + y_1 & & \\\\ & \\ddots & \\\\ & & y_n^2 + y_n \\end{bmatrix} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere we cannot solve for the matrix containing the $y$ variables because we cannot eliminate $\\mathbf{x}$ from the left side of the equation.\n\n- My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$ . Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$ . Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$ . My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$ . Commented Mar 31, 2021 at 5:45\n- 1 @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck! – Ralff Commented Mar 31, 2021 at 6:15\n- Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on. Commented Mar 31, 2021 at 17:38\n- @RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result. – Ralff Commented Mar 31, 2021 at 18:47\n\n## You must log in to answer this question.\n\n## Not the answer you're looking for? Browse other questions tagged\n\n## .\n", "main_html": "
where $\\mathbf{x}$, $\\mathbf{y}$, $\\mathbf{z}$, and $\\mathbf{u}$ are $1\\times n$ vectors and $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$ are diagonal $n\\times n$ matrices with $\\mathbf{x}$, $\\mathbf{y}$, and $\\mathbf{z}$, respectively, along their diagonals (i.e., $D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})$).
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My question is whether it is possible to solve for $\\mathbf{y}$ here – both the $\\mathbf{y}$ as a vector and the $\\mathbf{y}$ along the diagonal of $D_{\\mathbf{y}}$. The problem is that I do not know of operations to pull $\\mathbf{y}$ out of $D_{\\mathbf{y}}$. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
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$$(\\mathbf{y}e^T) \\odot I_n = D_{\\mathbf{y}}$$
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where $\\odot$ is the Hadamard product and $e^T = (1,1,\\ldots)\\in\\mathbb R^n$. So, would it be possible to distribute out $\\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\\mathbf{y}$, not $D_{\\mathbf{y}}$:
I think that in this case, it would not be possible to solve for a single vector $\\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:
where $\\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\\mathbf{y}$ – that is, $\\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}$.
where $\\mathbf{M}$ is an $n \\times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\\mathbf{yM}}$, we have some linear combination.
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I think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\\mathbf{M}$ and $D_{\\mathbf{x}}$ that is multiplied by a scalar ($n$, I believe). Then, we would have a set of vectors ($n-1$, I believe) that each have as elements different products of the elements in $\\mathbf{y}$ (e.g., $y_1 y_3$) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\\mathbf{M}$ and $D_{\\mathbf{x}}$.
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So, I have two questions here:
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Is my general intuition about what this equation would look like correct?
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Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?
I am not sure what the notation $D_\\mathbf{x}$, $D_\\mathbf{y}$, and $D_\\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\\mathbf{x}}$ is invertible, you may solve for $\\mathbf{y}$ as follows given the original equation:
Note: I can't verify that your edit is correct because I don't understand what $D_{\\mathbf{x} + \\mathbf{z}}$ represents.
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Edit 1:
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Thank you for the response as I didn't realize the meaning of the matrices $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
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Since $D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})$ and similarly for the other matrices, we have
As for the case where you have $\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}$ where $\\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}$, you would not be able to solve for $\\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
where we cannot solve for the matrix containing the $y$ variables because we cannot eliminate $\\mathbf{x}$ from the left side of the equation.
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\r\n $\\begingroup$\r\n My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$.\r\n $\\endgroup$\r\n \r\n \r\n Commented\r\n Mar 31, 2021 at 5:45\r\n \r\n \r\n \r\n \r\n
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\r\n $\\begingroup$\r\n @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck!\r\n $\\endgroup$\r\n
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\r\n $\\begingroup$\r\n Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on.\r\n $\\endgroup$\r\n \r\n \r\n Commented\r\n Mar 31, 2021 at 17:38\r\n \r\n
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\r\n $\\begingroup$\r\n @RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result.\r\n $\\endgroup$\r\n
where \\mathbf{x}, \\mathbf{y}, \\mathbf{z}, and \\mathbf{u} are 1\\times n vectors and D_{\\mathbf{x}}, D_{\\mathbf{y}}, and D_{\\mathbf{z}} are diagonal n\\times n matrices with \\mathbf{x}, \\mathbf{y}, and \\mathbf{z}, respectively, along their diagonals (i.e., D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})).
My question is whether it is possible to solve for \\mathbf{y} here – both the \\mathbf{y} as a vector and the \\mathbf{y} along the diagonal of D_{\\mathbf{y}}. The problem is that I do not know of operations to pull \\mathbf{y} out of D_{\\mathbf{y}}. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute \\mathbf{y} out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
", "content": [{"c": "My question is whether it is possible to solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "here – both the", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as a vector and the", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "along the diagonal of", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ". The problem is that I do not know of operations to pull", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "out of", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ". The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where \\odot is the Hadamard product and e^T = (1,1,\\ldots)\\in\\mathbb R^n. So, would it be possible to distribute out \\mathbf{y} and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to \\mathbf{y}, not D_{\\mathbf{y}}:
", "content": [{"c": "where", "t": "text"}, {"c": "\\odot", "t": "equation-inline"}, {"c": "is the Hadamard product and", "t": "text"}, {"c": "e^T = (1,1,\\ldots)\\in\\mathbb R^n", "t": "equation-inline"}, {"c": ". So, would it be possible to distribute out", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": ", not", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ":", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
Edit: Oh, it seems that extracting \\mathbf{y} in this case would be a simple as rewriting the equation above as
", "content": [{"c": "Edit: Oh, it seems that extracting", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "in this case would be a simple as rewriting the equation above as", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
because rewriting the equation this way would not change the terms along the diagonal of \\mathbf{z}D_{\\mathbf{y}}. Then we can write
", "content": [{"c": "because rewriting the equation this way would not change the terms along the diagonal of", "t": "text"}, {"c": "\\mathbf{z}D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ". Then we can write", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
But then how would solve for \\mathbf{y} in the following?
", "content": [{"c": "But then how would solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "in the following?", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
I think that in this case, it would not be possible to solve for a single vector \\mathbf{y} as in the previous case. Instead, we would have a system of polynomials:
", "content": [{"c": "I think that in this case, it would not be possible to solve for a single vector", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as in the previous case. Instead, we would have a system of polynomials:", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where \\mathbf{y^2} is a vector where the elements are the squares of the corresponding elements of \\mathbf{y} – that is, $\\mathbf{y^2} =
", "content": [{"c": "where", "t": "text"}, {"c": "\\mathbf{y^2}", "t": "equation-inline"}, {"c": "is a vector where the elements are the squares of the corresponding elements of", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "– that is,$\\mathbf{y^2} =", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where \\mathbf{M} is an n \\times n matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of D_{\\mathbf{yM}}, we have some linear combination.
", "content": [{"c": "where", "t": "text"}, {"c": "\\mathbf{M}", "t": "equation-inline"}, {"c": "is an", "t": "text"}, {"c": "n \\times n", "t": "equation-inline"}, {"c": "matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of", "t": "text"}, {"c": "D_{\\mathbf{yM}}", "t": "equation-inline"}, {"c": ", we have some linear combination.", "t": "text"}]}, {"type": "paragraph", "raw_content": "
I think in this case, the polynomial system of equations would be rather complicated. We would have the vector \\mathbf{y^2} again, but this time multiplied by some diagonal matrix based on the elements of \\mathbf{M} and D_{\\mathbf{x}} that is multiplied by a scalar (n, I believe). Then, we would have a set of vectors (n-1, I believe) that each have as elements different products of the elements in \\mathbf{y} (e.g., y_1 y_3) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in \\mathbf{M} and D_{\\mathbf{x}}.
", "content": [{"c": "I think in this case, the polynomial system of equations would be rather complicated. We would have the vector", "t": "text"}, {"c": "\\mathbf{y^2}", "t": "equation-inline"}, {"c": "again, but this time multiplied by some diagonal matrix based on the elements of", "t": "text"}, {"c": "\\mathbf{M}", "t": "equation-inline"}, {"c": "and", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": "that is multiplied by a scalar (", "t": "text"}, {"c": "n", "t": "equation-inline"}, {"c": ", I believe). Then, we would have a set of vectors (", "t": "text"}, {"c": "n-1", "t": "equation-inline"}, {"c": ", I believe) that each have as elements different products of the elements in", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "(e.g.,", "t": "text"}, {"c": "y_1 y_3", "t": "equation-inline"}, {"c": ") and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in", "t": "text"}, {"c": "\\mathbf{M}", "t": "equation-inline"}, {"c": "and", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": ".", "t": "text"}]}, {"type": "paragraph", "raw_content": "
So, I have two questions here:
", "content": [{"c": "So, I have two questions here:", "t": "text"}]}, {"type": "list", "raw_content": "
Is my general intuition about what this equation would look like correct?
Are there techniques to solve for the elements of \\mathbf{y} in this system?
", "content": {"items": [{"c": "Is my general intuition about what this equation would look like correct?"}, {"c": "Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?"}], "list_attribute": "ordered", "list_nest_level": "1"}}, {"type": "title", "raw_content": "
I am not sure what the notation D_\\mathbf{x}, D_\\mathbf{y}, and D_\\mathbf{z} represent and if the importance is significant, but assuming D_{\\mathbf{x}} is invertible, you may solve for \\mathbf{y} as follows given the original equation:
", "content": [{"c": "I am not sure what the notation", "t": "text"}, {"c": "D_\\mathbf{x}", "t": "equation-inline"}, {"c": ",", "t": "text"}, {"c": "D_\\mathbf{y}", "t": "equation-inline"}, {"c": ", and", "t": "text"}, {"c": "D_\\mathbf{z}", "t": "equation-inline"}, {"c": "represent and if the importance is significant, but assuming", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": "is invertible, you may solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as follows given the original equation:", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
Thank you for the response as I didn't realize the meaning of the matrices D_{\\mathbf{x}}, D_{\\mathbf{y}}, and D_{\\mathbf{z}}. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
", "content": [{"c": "Thank you for the response as I didn't realize the meaning of the matrices", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": ",", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ", and", "t": "text"}, {"c": "D_{\\mathbf{z}}", "t": "equation-inline"}, {"c": ". Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.", "t": "text"}]}, {"type": "paragraph", "raw_content": "
Since D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x}) and similarly for the other matrices, we have
", "content": [{"c": "Since", "t": "text"}, {"c": "D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})", "t": "equation-inline"}, {"c": "and similarly for the other matrices, we have", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
$, you would not be able to solve for \\mathbf{y} as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
", "content": [{"c": "$, you would not be able to solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where we cannot solve for the matrix containing the y variables because we cannot eliminate \\mathbf{x} from the left side of the equation.
", "content": [{"c": "where we cannot solve for the matrix containing the", "t": "text"}, {"c": "y", "t": "equation-inline"}, {"c": "variables because we cannot eliminate", "t": "text"}, {"c": "\\mathbf{x}", "t": "equation-inline"}, {"c": "from the left side of the equation.", "t": "text"}]}, {"type": "list", "raw_content": "
My notation for, for example, D_{\\mathbf{x}} was intended to refer to \\mathrm{diag}(\\mathbf{x}). Thus, D_{\\mathbf{x+y}} means \\mathrm{diag}(\\mathbf{x+y}). Sorry if that was unclear. Furthermore, in the solution you provided, \\mathbf{y} is still along the diagonal of D_{\\mathbf{y}}. My aim is to remove it so we can have an non-implicit expression for \\mathbf{y}.CommentedMar 31, 2021 at 5:45
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@RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck!
Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on.CommentedMar 31, 2021 at 17:38
@RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the \\mathbf{y}^2 case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result.
", "content": {"items": [{"c": "My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$ . Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$ . Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$ . My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$ . Commented Mar 31, 2021 at 5:45"}, {"c": "1 @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck! – Ralff Commented Mar 31, 2021 at 6:15"}, {"c": "Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on. Commented Mar 31, 2021 at 17:38"}, {"c": "@RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result. – Ralff Commented Mar 31, 2021 at 18:47"}], "list_attribute": "unordered", "list_nest_level": "1"}}, {"type": "title", "raw_content": "
\r\n\t\t\t\t\t\tStack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.\r\n\t\t\t\t\t
where $\\mathbf{x}$, $\\mathbf{y}$, $\\mathbf{z}$, and $\\mathbf{u}$ are $1\\times n$ vectors and $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$ are diagonal $n\\times n$ matrices with $\\mathbf{x}$, $\\mathbf{y}$, and $\\mathbf{z}$, respectively, along their diagonals (i.e., $D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})$).
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My question is whether it is possible to solve for $\\mathbf{y}$ here – both the $\\mathbf{y}$ as a vector and the $\\mathbf{y}$ along the diagonal of $D_{\\mathbf{y}}$. The problem is that I do not know of operations to pull $\\mathbf{y}$ out of $D_{\\mathbf{y}}$. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
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$$(\\mathbf{y}e^T) \\odot I_n = D_{\\mathbf{y}}$$
\r\n
where $\\odot$ is the Hadamard product and $e^T = (1,1,\\ldots)\\in\\mathbb R^n$. So, would it be possible to distribute out $\\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\\mathbf{y}$, not $D_{\\mathbf{y}}$:
I think that in this case, it would not be possible to solve for a single vector $\\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:
where $\\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\\mathbf{y}$ – that is, $\\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}$.
where $\\mathbf{M}$ is an $n \\times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\\mathbf{yM}}$, we have some linear combination.
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I think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\\mathbf{M}$ and $D_{\\mathbf{x}}$ that is multiplied by a scalar ($n$, I believe). Then, we would have a set of vectors ($n-1$, I believe) that each have as elements different products of the elements in $\\mathbf{y}$ (e.g., $y_1 y_3$) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\\mathbf{M}$ and $D_{\\mathbf{x}}$.
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So, I have two questions here:
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Is my general intuition about what this equation would look like correct?
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Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?
I am not sure what the notation $D_\\mathbf{x}$, $D_\\mathbf{y}$, and $D_\\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\\mathbf{x}}$ is invertible, you may solve for $\\mathbf{y}$ as follows given the original equation:
Note: I can't verify that your edit is correct because I don't understand what $D_{\\mathbf{x} + \\mathbf{z}}$ represents.
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Edit 1:
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Thank you for the response as I didn't realize the meaning of the matrices $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
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Since $D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})$ and similarly for the other matrices, we have
As for the case where you have $\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}$ where $\\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}$, you would not be able to solve for $\\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
\r\n $\\begingroup$\r\n My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$.\r\n $\\endgroup$\r\n
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\r\n $\\begingroup$\r\n @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck!\r\n $\\endgroup$\r\n
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\r\n $\\begingroup$\r\n Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on.\r\n $\\endgroup$\r\n
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\r\n $\\begingroup$\r\n @RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result.\r\n $\\endgroup$\r\n
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+{"url": "https://math.stackexchange.com/questions/4082284/solving-for-vector-contained-in-a-diagonal-matrix", "content": "Consider the following system of equations:\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{z}D_{\\mathbf{y}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere $\\mathbf{x}$ , $\\mathbf{y}$ , $\\mathbf{z}$ , and $\\mathbf{u}$ are $1\\times n$ vectors and $D_{\\mathbf{x}}$ , $D_{\\mathbf{y}}$ , and $D_{\\mathbf{z}}$ are diagonal $n\\times n$ matrices with $\\mathbf{x}$ , $\\mathbf{y}$ , and $\\mathbf{z}$ , respectively, along their diagonals (i.e., $D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})$ ).\n\nMy question is whether it is possible to solve for $\\mathbf{y}$ here – both the $\\mathbf{y}$ as a vector and the $\\mathbf{y}$ along the diagonal of $D_{\\mathbf{y}}$ . The problem is that I do not know of operations to pull $\\mathbf{y}$ out of $D_{\\mathbf{y}}$ . The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write\n\n$$\n(\\mathbf{y}e^T) \\odot I_n = D_{\\mathbf{y}}\n$$\n\nwhere $\\odot$ is the Hadamard product and $e^T = (1,1,\\ldots)\\in\\mathbb R^n$ . So, would it be possible to distribute out $\\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\\mathbf{y}$ , not $D_{\\mathbf{y}}$ :\n\n$$\n\\mathbf{y}(e^T \\odot I_n) \\neq D_{\\mathbf{y}}\n$$\n\nEdit: Oh, it seems that extracting $\\mathbf{y}$ in this case would be a simple as rewriting the equation above as\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{y}D_{\\mathbf{z}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nbecause rewriting the equation this way would not change the terms along the diagonal of $\\mathbf{z}D_{\\mathbf{y}}$ . Then we can write\n\n$$\n\\mathbf{y}=\\mathbf{u}D_{\\mathbf{z}}D_{\\mathbf{x+z}}^{-1}\n$$\n\nBut then how would solve for $\\mathbf{y}$ in the following?\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{y}D_{\\mathbf{y}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nI think that in this case, it would not be possible to solve for a single vector $\\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:\n\n$$\n\\mathbf{y^2} + \\mathbf{y}D_{\\mathbf{x}} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere $\\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\\mathbf{y}$ – that is, $\\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}$ .\n\nNext, what if we had an equation as follows?\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{y}D_{\\mathbf{yM}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere $\\mathbf{M}$ is an $n \\times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\\mathbf{yM}}$ , we have some linear combination.\n\nI think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\\mathbf{M}$ and $D_{\\mathbf{x}}$ that is multiplied by a scalar ( $n$ , I believe). Then, we would have a set of vectors ( $n-1$ , I believe) that each have as elements different products of the elements in $\\mathbf{y}$ (e.g., $y_1 y_3$ ) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\\mathbf{M}$ and $D_{\\mathbf{x}}$ .\n\nSo, I have two questions here:\n\n1. Is my general intuition about what this equation would look like correct?\n2. Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?\n\n## 1 Answer\n\n1\n\nI am not sure what the notation $D_\\mathbf{x}$ , $D_\\mathbf{y}$ , and $D_\\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\\mathbf{x}}$ is invertible, you may solve for $\\mathbf{y}$ as follows given the original equation:\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}+\\mathbf{z}D_{\\mathbf{y}}=\\mathbf{u}D_{\\mathbf{z}}\n$$\n\n$$\n\\mathbf{y}D_{\\mathbf{x}}=\\mathbf{u}D_{\\mathbf{z}}-\\mathbf{z}D_{\\mathbf{y}}\n$$\n\n$$\n\\mathbf{y}=\\bigr( \\mathbf{u}D_{\\mathbf{z}}-\\mathbf{z}D_{\\mathbf{y}} \\bigr) D_{\\mathbf{x}}^{-1}\n$$\n\nNote: I can't verify that your edit is correct because I don't understand what $D_{\\mathbf{x} + \\mathbf{z}}$ represents.\n\nEdit 1:\n\nThank you for the response as I didn't realize the meaning of the matrices $D_{\\mathbf{x}}$ , $D_{\\mathbf{y}}$ , and $D_{\\mathbf{z}}$ . Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.\n\nSince $D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})$ and similarly for the other matrices, we have\n\n$$\n\\begin{bmatrix} y_1 & \\cdots & y_n \\end{bmatrix} \\begin{bmatrix} x_1 & & \\\\ & \\ddots & \\\\ & & x_n\\end{bmatrix} + \\begin{bmatrix} z_1 & \\cdots & z_n \\end{bmatrix} \\begin{bmatrix} y_1 & & \\\\ & \\ddots & \\\\ & & y_n\\end{bmatrix} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nand multiplying through we have\n\n$$\n\\begin{align}\n\\begin{bmatrix} y_1 x_1 & \\cdots & y_n x_n \\end{bmatrix} + \\begin{bmatrix} y_1 z_1 & \\cdots & y_n z_n \\end{bmatrix} &= \\mathbf{u}D_{\\mathbf{z}}\\\\\n\\begin{bmatrix} y_1 x_1 + y_1 z_1 & \\cdots & y_n x_n + y_n z_n \\end{bmatrix} &= \\mathbf{u}D_{\\mathbf{z}} \\\\\n\\begin{bmatrix} y_1 (x_1 + z_1) & \\cdots & y_n(x_n + z_n) \\end{bmatrix} &= \\mathbf{u}D_{\\mathbf{z}} \\\\\n\\end{align}\n$$\n\nTherefore, this can be written in matrix form as\n\n$$\n\\begin{bmatrix} y_1 & \\cdots & y_n \\end{bmatrix} \\begin{bmatrix} x_1 + z_1 & & \\\\ & \\ddots & \\\\ & & x_n + z_n \\end{bmatrix} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nor more concisely as\n\n$$\n\\mathbf{y} (D_{\\mathbf{x}} + D_{\\mathbf{z}} ) = \\mathbf{u} D_{\\mathbf{z}}\n$$\n\nwhich is exactly what is given in your edit:\n\n$$\n\\mathbf{y} = \\mathbf{u} D_{\\mathbf{z}} (D_{\\mathbf{x}} + D_{\\mathbf{z}} ) ^{-1}\n$$\n\nEdit 2:\n\nAs for the case where you have $\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}$ where $\\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}$ , you would not be able to solve for $\\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):\n\n$$\n\\begin{bmatrix} x_1 & \\cdots & x_n \\end{bmatrix} \\begin{bmatrix} y_1^2 + y_1 & & \\\\ & \\ddots & \\\\ & & y_n^2 + y_n \\end{bmatrix} = \\mathbf{u}D_{\\mathbf{z}}\n$$\n\nwhere we cannot solve for the matrix containing the $y$ variables because we cannot eliminate $\\mathbf{x}$ from the left side of the equation.\n\n- My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$. Commented Mar 31, 2021 at 5:45\n- 1 @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck! – Ralff Commented Mar 31, 2021 at 6:15\n- Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on. Commented Mar 31, 2021 at 17:38\n- @RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result. – Ralff Commented Mar 31, 2021 at 18:47\n\n## You must log in to answer this question.\n\n## Not the answer you're looking for? Browse other questions tagged\n\n## .\n", "main_html": "
where $\\mathbf{x}$, $\\mathbf{y}$, $\\mathbf{z}$, and $\\mathbf{u}$ are $1\\times n$ vectors and $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$ are diagonal $n\\times n$ matrices with $\\mathbf{x}$, $\\mathbf{y}$, and $\\mathbf{z}$, respectively, along their diagonals (i.e., $D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})$).
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My question is whether it is possible to solve for $\\mathbf{y}$ here – both the $\\mathbf{y}$ as a vector and the $\\mathbf{y}$ along the diagonal of $D_{\\mathbf{y}}$. The problem is that I do not know of operations to pull $\\mathbf{y}$ out of $D_{\\mathbf{y}}$. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
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$$(\\mathbf{y}e^T) \\odot I_n = D_{\\mathbf{y}}$$
\r\n
where $\\odot$ is the Hadamard product and $e^T = (1,1,\\ldots)\\in\\mathbb R^n$. So, would it be possible to distribute out $\\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\\mathbf{y}$, not $D_{\\mathbf{y}}$:
I think that in this case, it would not be possible to solve for a single vector $\\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:
where $\\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\\mathbf{y}$ – that is, $\\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}$.
where $\\mathbf{M}$ is an $n \\times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\\mathbf{yM}}$, we have some linear combination.
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I think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\\mathbf{M}$ and $D_{\\mathbf{x}}$ that is multiplied by a scalar ($n$, I believe). Then, we would have a set of vectors ($n-1$, I believe) that each have as elements different products of the elements in $\\mathbf{y}$ (e.g., $y_1 y_3$) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\\mathbf{M}$ and $D_{\\mathbf{x}}$.
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So, I have two questions here:
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Is my general intuition about what this equation would look like correct?
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Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?
I am not sure what the notation $D_\\mathbf{x}$, $D_\\mathbf{y}$, and $D_\\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\\mathbf{x}}$ is invertible, you may solve for $\\mathbf{y}$ as follows given the original equation:
Note: I can't verify that your edit is correct because I don't understand what $D_{\\mathbf{x} + \\mathbf{z}}$ represents.
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Edit 1:
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Thank you for the response as I didn't realize the meaning of the matrices $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
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Since $D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})$ and similarly for the other matrices, we have
As for the case where you have $\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}$ where $\\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}$, you would not be able to solve for $\\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
where we cannot solve for the matrix containing the $y$ variables because we cannot eliminate $\\mathbf{x}$ from the left side of the equation.
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\r\n $\\begingroup$\r\n My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$.\r\n $\\endgroup$\r\n \r\n \r\n Commented\r\n Mar 31, 2021 at 5:45\r\n \r\n \r\n \r\n \r\n
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\r\n $\\begingroup$\r\n @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck!\r\n $\\endgroup$\r\n
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\r\n $\\begingroup$\r\n Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on.\r\n $\\endgroup$\r\n \r\n \r\n Commented\r\n Mar 31, 2021 at 17:38\r\n \r\n
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\r\n $\\begingroup$\r\n @RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result.\r\n $\\endgroup$\r\n
where \\mathbf{x}, \\mathbf{y}, \\mathbf{z}, and \\mathbf{u} are 1\\times n vectors and D_{\\mathbf{x}}, D_{\\mathbf{y}}, and D_{\\mathbf{z}} are diagonal n\\times n matrices with \\mathbf{x}, \\mathbf{y}, and \\mathbf{z}, respectively, along their diagonals (i.e., D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})).
My question is whether it is possible to solve for \\mathbf{y} here – both the \\mathbf{y} as a vector and the \\mathbf{y} along the diagonal of D_{\\mathbf{y}}. The problem is that I do not know of operations to pull \\mathbf{y} out of D_{\\mathbf{y}}. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute \\mathbf{y} out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
", "content": [{"c": "My question is whether it is possible to solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "here – both the", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as a vector and the", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "along the diagonal of", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ". The problem is that I do not know of operations to pull", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "out of", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ". The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where \\odot is the Hadamard product and e^T = (1,1,\\ldots)\\in\\mathbb R^n. So, would it be possible to distribute out \\mathbf{y} and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to \\mathbf{y}, not D_{\\mathbf{y}}:
", "content": [{"c": "where", "t": "text"}, {"c": "\\odot", "t": "equation-inline"}, {"c": "is the Hadamard product and", "t": "text"}, {"c": "e^T = (1,1,\\ldots)\\in\\mathbb R^n", "t": "equation-inline"}, {"c": ". So, would it be possible to distribute out", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": ", not", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ":", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
Edit: Oh, it seems that extracting \\mathbf{y} in this case would be a simple as rewriting the equation above as
", "content": [{"c": "Edit: Oh, it seems that extracting", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "in this case would be a simple as rewriting the equation above as", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
because rewriting the equation this way would not change the terms along the diagonal of \\mathbf{z}D_{\\mathbf{y}}. Then we can write
", "content": [{"c": "because rewriting the equation this way would not change the terms along the diagonal of", "t": "text"}, {"c": "\\mathbf{z}D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ". Then we can write", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
But then how would solve for \\mathbf{y} in the following?
", "content": [{"c": "But then how would solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "in the following?", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
I think that in this case, it would not be possible to solve for a single vector \\mathbf{y} as in the previous case. Instead, we would have a system of polynomials:
", "content": [{"c": "I think that in this case, it would not be possible to solve for a single vector", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as in the previous case. Instead, we would have a system of polynomials:", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where \\mathbf{y^2} is a vector where the elements are the squares of the corresponding elements of \\mathbf{y} – that is, \\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}.
", "content": [{"c": "where", "t": "text"}, {"c": "\\mathbf{y^2}", "t": "equation-inline"}, {"c": "is a vector where the elements are the squares of the corresponding elements of", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "– that is,", "t": "text"}, {"c": "\\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}", "t": "equation-inline"}, {"c": ".", "t": "text"}]}, {"type": "paragraph", "raw_content": "
Next, what if we had an equation as follows?
", "content": [{"c": "Next, what if we had an equation as follows?", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where \\mathbf{M} is an n \\times n matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of D_{\\mathbf{yM}}, we have some linear combination.
", "content": [{"c": "where", "t": "text"}, {"c": "\\mathbf{M}", "t": "equation-inline"}, {"c": "is an", "t": "text"}, {"c": "n \\times n", "t": "equation-inline"}, {"c": "matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of", "t": "text"}, {"c": "D_{\\mathbf{yM}}", "t": "equation-inline"}, {"c": ", we have some linear combination.", "t": "text"}]}, {"type": "paragraph", "raw_content": "
I think in this case, the polynomial system of equations would be rather complicated. We would have the vector \\mathbf{y^2} again, but this time multiplied by some diagonal matrix based on the elements of \\mathbf{M} and D_{\\mathbf{x}} that is multiplied by a scalar (n, I believe). Then, we would have a set of vectors (n-1, I believe) that each have as elements different products of the elements in \\mathbf{y} (e.g., y_1 y_3) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in \\mathbf{M} and D_{\\mathbf{x}}.
", "content": [{"c": "I think in this case, the polynomial system of equations would be rather complicated. We would have the vector", "t": "text"}, {"c": "\\mathbf{y^2}", "t": "equation-inline"}, {"c": "again, but this time multiplied by some diagonal matrix based on the elements of", "t": "text"}, {"c": "\\mathbf{M}", "t": "equation-inline"}, {"c": "and", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": "that is multiplied by a scalar (", "t": "text"}, {"c": "n", "t": "equation-inline"}, {"c": ", I believe). Then, we would have a set of vectors (", "t": "text"}, {"c": "n-1", "t": "equation-inline"}, {"c": ", I believe) that each have as elements different products of the elements in", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "(e.g.,", "t": "text"}, {"c": "y_1 y_3", "t": "equation-inline"}, {"c": ") and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in", "t": "text"}, {"c": "\\mathbf{M}", "t": "equation-inline"}, {"c": "and", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": ".", "t": "text"}]}, {"type": "paragraph", "raw_content": "
So, I have two questions here:
", "content": [{"c": "So, I have two questions here:", "t": "text"}]}, {"type": "list", "raw_content": "
Is my general intuition about what this equation would look like correct?
Are there techniques to solve for the elements of \\mathbf{y} in this system?
", "content": {"items": [{"c": "Is my general intuition about what this equation would look like correct?"}, {"c": "Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?"}], "list_attribute": "ordered", "list_nest_level": "1"}}, {"type": "title", "raw_content": "
I am not sure what the notation D_\\mathbf{x}, D_\\mathbf{y}, and D_\\mathbf{z} represent and if the importance is significant, but assuming D_{\\mathbf{x}} is invertible, you may solve for \\mathbf{y} as follows given the original equation:
", "content": [{"c": "I am not sure what the notation", "t": "text"}, {"c": "D_\\mathbf{x}", "t": "equation-inline"}, {"c": ",", "t": "text"}, {"c": "D_\\mathbf{y}", "t": "equation-inline"}, {"c": ", and", "t": "text"}, {"c": "D_\\mathbf{z}", "t": "equation-inline"}, {"c": "represent and if the importance is significant, but assuming", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": "is invertible, you may solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as follows given the original equation:", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
Thank you for the response as I didn't realize the meaning of the matrices D_{\\mathbf{x}}, D_{\\mathbf{y}}, and D_{\\mathbf{z}}. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
", "content": [{"c": "Thank you for the response as I didn't realize the meaning of the matrices", "t": "text"}, {"c": "D_{\\mathbf{x}}", "t": "equation-inline"}, {"c": ",", "t": "text"}, {"c": "D_{\\mathbf{y}}", "t": "equation-inline"}, {"c": ", and", "t": "text"}, {"c": "D_{\\mathbf{z}}", "t": "equation-inline"}, {"c": ". Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.", "t": "text"}]}, {"type": "paragraph", "raw_content": "
Since D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x}) and similarly for the other matrices, we have
", "content": [{"c": "Since", "t": "text"}, {"c": "D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})", "t": "equation-inline"}, {"c": "and similarly for the other matrices, we have", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
As for the case where you have \\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}} where \\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}, you would not be able to solve for \\mathbf{y} as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
", "content": [{"c": "As for the case where you have", "t": "text"}, {"c": "\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}", "t": "equation-inline"}, {"c": "where", "t": "text"}, {"c": "\\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}", "t": "equation-inline"}, {"c": ", you would not be able to solve for", "t": "text"}, {"c": "\\mathbf{y}", "t": "equation-inline"}, {"c": "as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):", "t": "text"}]}, {"type": "equation-interline", "raw_content": "
where we cannot solve for the matrix containing the y variables because we cannot eliminate \\mathbf{x} from the left side of the equation.
", "content": [{"c": "where we cannot solve for the matrix containing the", "t": "text"}, {"c": "y", "t": "equation-inline"}, {"c": "variables because we cannot eliminate", "t": "text"}, {"c": "\\mathbf{x}", "t": "equation-inline"}, {"c": "from the left side of the equation.", "t": "text"}]}, {"type": "list", "raw_content": "
My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$.CommentedMar 31, 2021 at 5:45
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@RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck!
Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on.CommentedMar 31, 2021 at 17:38
@RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result.
", "content": {"items": [{"c": "My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$. Commented Mar 31, 2021 at 5:45"}, {"c": "1 @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck! – Ralff Commented Mar 31, 2021 at 6:15"}, {"c": "Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on. Commented Mar 31, 2021 at 17:38"}, {"c": "@RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result. – Ralff Commented Mar 31, 2021 at 18:47"}], "list_attribute": "unordered", "list_nest_level": "1"}}, {"type": "title", "raw_content": "
\r\n\t\t\t\t\t\tStack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.\r\n\t\t\t\t\t
where $\\mathbf{x}$, $\\mathbf{y}$, $\\mathbf{z}$, and $\\mathbf{u}$ are $1\\times n$ vectors and $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$ are diagonal $n\\times n$ matrices with $\\mathbf{x}$, $\\mathbf{y}$, and $\\mathbf{z}$, respectively, along their diagonals (i.e., $D_{\\mathbf{x}} = \\mathrm{diag}(\\mathbf{x})$).
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My question is whether it is possible to solve for $\\mathbf{y}$ here – both the $\\mathbf{y}$ as a vector and the $\\mathbf{y}$ along the diagonal of $D_{\\mathbf{y}}$. The problem is that I do not know of operations to pull $\\mathbf{y}$ out of $D_{\\mathbf{y}}$. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
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$$(\\mathbf{y}e^T) \\odot I_n = D_{\\mathbf{y}}$$
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where $\\odot$ is the Hadamard product and $e^T = (1,1,\\ldots)\\in\\mathbb R^n$. So, would it be possible to distribute out $\\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\\mathbf{y}$, not $D_{\\mathbf{y}}$:
I think that in this case, it would not be possible to solve for a single vector $\\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:
where $\\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\\mathbf{y}$ – that is, $\\mathbf{y^2} = \\begin{pmatrix} y_{1}^2 & y_{2}^2 & \\cdots & y_{n}^2 \\end{pmatrix}$.
where $\\mathbf{M}$ is an $n \\times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\\mathbf{yM}}$, we have some linear combination.
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I think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\\mathbf{M}$ and $D_{\\mathbf{x}}$ that is multiplied by a scalar ($n$, I believe). Then, we would have a set of vectors ($n-1$, I believe) that each have as elements different products of the elements in $\\mathbf{y}$ (e.g., $y_1 y_3$) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\\mathbf{M}$ and $D_{\\mathbf{x}}$.
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So, I have two questions here:
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Is my general intuition about what this equation would look like correct?
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Are there techniques to solve for the elements of $\\mathbf{y}$ in this system?
I am not sure what the notation $D_\\mathbf{x}$, $D_\\mathbf{y}$, and $D_\\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\\mathbf{x}}$ is invertible, you may solve for $\\mathbf{y}$ as follows given the original equation:
Note: I can't verify that your edit is correct because I don't understand what $D_{\\mathbf{x} + \\mathbf{z}}$ represents.
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Edit 1:
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Thank you for the response as I didn't realize the meaning of the matrices $D_{\\mathbf{x}}$, $D_{\\mathbf{y}}$, and $D_{\\mathbf{z}}$. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
\r\n
Since $D_{\\mathbf{x}} = \\text{diag}(\\mathbf{x})$ and similarly for the other matrices, we have
As for the case where you have $\\mathbf{y}^2 + \\mathbf{y} D_{\\mathbf{x}} = \\mathbf{u} D_{\\mathbf{z}}$ where $\\mathbf{y}^2 = \\begin{bmatrix} y_1^2 & \\cdots & y_n^2 \\end{bmatrix}$, you would not be able to solve for $\\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
\r\n $\\begingroup$\r\n My notation for, for example, $D_{\\mathbf{x}}$ was intended to refer to $\\mathrm{diag}(\\mathbf{x})$. Thus, $D_{\\mathbf{x+y}}$ means $\\mathrm{diag}(\\mathbf{x+y})$. Sorry if that was unclear. Furthermore, in the solution you provided, $\\mathbf{y}$ is still along the diagonal of $D_{\\mathbf{y}}$. My aim is to remove it so we can have an non-implicit expression for $\\mathbf{y}$.\r\n $\\endgroup$\r\n
\r\n \r\n Commented\r\n Mar 31, 2021 at 5:45\r\n \r\n \r\n \r\n \r\n
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\r\n $\\begingroup$\r\n @RyandaSilva Ah. Thanks. I apologize I misunderstood the notation. I edited my question of how I would work the problem if that may be of any use to you. Your edit is indeed correct. Best of luck!\r\n $\\endgroup$\r\n
\r\n \r\n Commented\r\n Mar 31, 2021 at 6:15\r\n \r\n \r\n \r\n \r\n
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\r\n $\\begingroup$\r\n Thank you very much for the help and the kind wishes. If you have time, I have added a bit more to my original post that I would love to get your feedback on.\r\n $\\endgroup$\r\n
\r\n \r\n Commented\r\n Mar 31, 2021 at 17:38\r\n \r\n
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\r\n $\\begingroup$\r\n @RyandaSilva You're welcome! I made a mistake, so I updated my answer. Please, see the changes. I provided a hint for your additional edit but only for the $\\mathbf{y}^2$ case. If you have a matrix with off diagonal terms, then the matrix multiplication will be slightly more complicated, but I suggest working out the multiplication by hand, so you can see the result.\r\n $\\endgroup$\r\n
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