Mixing
Singular Value Decomposition of a Real Unit Matrix
Given a real matrix $A in mathbb{R}^{m times n}$ whose entries are all ones, what is the reduced/short/economy singular value decomposition $A = USigma V^T$? I can see that we have a single singular value for $A$, namely $sqrt{mn}$, but I'm having trouble coming up with general formulas for the orthogonal matrices $U$ and $V$ in terms of $m$ and $n$.
linear-algebra matrices matrix-decomposition svd
asked Nov 20 '18 at 20:35
rcmpgrc
174
add a comment |
Given a real matrix $A in mathbb{R}^{m times n}$ whose entries are all ones, what is the reduced/short/economy singular value decomposition $A = USigma V^T$? I can see that we have a single singular value for $A$, namely $sqrt{mn}$, but I'm having trouble coming up with general formulas for the orthogonal matrices $U$ and $V$ in terms of $m$ and $n$.
linear-algebra matrices matrix-decomposition svd
asked Nov 20 '18 at 20:35
rcmpgrc
174
If you already know that there's a single singular value, then you also know that there's a single left singular vector and a single right singular vector, both of which are very easy to compute. Isn't that all the information you need for a reduced SVD?
– Qiaochu Yuan
Nov 21 '18 at 0:46
I thought for SVD of any given $m times n$ matrix there will be $m$ left singular vectors and $n$ right singular vectors? I've been trying to determine some kind of pattern by generating various unit matrices and using the svd function in MATLAB, but I can't discern anything reasonable.
– rcmpgrc
Nov 21 '18 at 3:14
Yes, that's true of SVD, which is non-unique. But only the singular vectors associated to the nonzero singular values matter; that's the point of the reduced SVD.
– Qiaochu Yuan
Nov 21 '18 at 4:25
add a comment |
Given a real matrix $A in mathbb{R}^{m times n}$ whose entries are all ones, what is the reduced/short/economy singular value decomposition $A = USigma V^T$? I can see that we have a single singular value for $A$, namely $sqrt{mn}$, but I'm having trouble coming up with general formulas for the orthogonal matrices $U$ and $V$ in terms of $m$ and $n$.
linear-algebra matrices matrix-decomposition svd
asked Nov 20 '18 at 20:35
rcmpgrc
174
Given a real matrix $A in mathbb{R}^{m times n}$ whose entries are all ones, what is the reduced/short/economy singular value decomposition $A = USigma V^T$? I can see that we have a single singular value for $A$, namely $sqrt{mn}$, but I'm having trouble coming up with general formulas for the orthogonal matrices $U$ and $V$ in terms of $m$ and $n$.
linear-algebra matrices matrix-decomposition svd
linear-algebra matrices matrix-decomposition svd
asked Nov 20 '18 at 20:35
rcmpgrc
174
asked Nov 20 '18 at 20:35
rcmpgrc
174
asked Nov 20 '18 at 20:35
rcmpgrc
174
asked Nov 20 '18 at 20:35
rcmpgrc
174
asked Nov 20 '18 at 20:35
rcmpgrc
174
174
If you already know that there's a single singular value, then you also know that there's a single left singular vector and a single right singular vector, both of which are very easy to compute. Isn't that all the information you need for a reduced SVD?
– Qiaochu Yuan
Nov 21 '18 at 0:46
I thought for SVD of any given $m times n$ matrix there will be $m$ left singular vectors and $n$ right singular vectors? I've been trying to determine some kind of pattern by generating various unit matrices and using the svd function in MATLAB, but I can't discern anything reasonable.
– rcmpgrc
Nov 21 '18 at 3:14
Yes, that's true of SVD, which is non-unique. But only the singular vectors associated to the nonzero singular values matter; that's the point of the reduced SVD.
– Qiaochu Yuan
Nov 21 '18 at 4:25
add a comment |
If you already know that there's a single singular value, then you also know that there's a single left singular vector and a single right singular vector, both of which are very easy to compute. Isn't that all the information you need for a reduced SVD?
– Qiaochu Yuan
Nov 21 '18 at 0:46
I thought for SVD of any given $m times n$ matrix there will be $m$ left singular vectors and $n$ right singular vectors? I've been trying to determine some kind of pattern by generating various unit matrices and using the svd function in MATLAB, but I can't discern anything reasonable.
– rcmpgrc
Nov 21 '18 at 3:14
Yes, that's true of SVD, which is non-unique. But only the singular vectors associated to the nonzero singular values matter; that's the point of the reduced SVD.
– Qiaochu Yuan
Nov 21 '18 at 4:25
If you already know that there's a single singular value, then you also know that there's a single left singular vector and a single right singular vector, both of which are very easy to compute. Isn't that all the information you need for a reduced SVD?
– Qiaochu Yuan
Nov 21 '18 at 0:46
If you already know that there's a single singular value, then you also know that there's a single left singular vector and a single right singular vector, both of which are very easy to compute. Isn't that all the information you need for a reduced SVD?
– Qiaochu Yuan
Nov 21 '18 at 0:46
I thought for SVD of any given $m times n$ matrix there will be $m$ left singular vectors and $n$ right singular vectors? I've been trying to determine some kind of pattern by generating various unit matrices and using the svd function in MATLAB, but I can't discern anything reasonable.
– rcmpgrc
Nov 21 '18 at 3:14
I thought for SVD of any given $m times n$ matrix there will be $m$ left singular vectors and $n$ right singular vectors? I've been trying to determine some kind of pattern by generating various unit matrices and using the svd function in MATLAB, but I can't discern anything reasonable.
– rcmpgrc
Nov 21 '18 at 3:14
Yes, that's true of SVD, which is non-unique. But only the singular vectors associated to the nonzero singular values matter; that's the point of the reduced SVD.
– Qiaochu Yuan
Nov 21 '18 at 4:25
Yes, that's true of SVD, which is non-unique. But only the singular vectors associated to the nonzero singular values matter; that's the point of the reduced SVD.
– Qiaochu Yuan
Nov 21 '18 at 4:25
add a comment |
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Suppose $A$ has $k$ nonzero singular values (i.e. $sigma_k>sigma_{k+1}=0$). Partition $U$ as $pmatrix{U_1&U_2}$ and $V$ as $pmatrix{V_1&V_2}$, where each of $U_1$ and $V_1$ has $k$ columns. The so-called reduced/economic SVD is given by $U_1operatorname{diag}(sigma_1,ldots,sigma_k)V_1^T$. Note that the result of this matrix product is precisely equal to $A=USV^T$. The terms "reduced" or "economic" refer not to the matrix product, but to the sizes of multiplicands: since $U_1,operatorname{diag}(sigma_1,ldots,sigma_k)$ and $V_1$ have smaller sizes than $U,Sigma$ and $V$, it is more economical to store them than to store a full SVD.
answered Nov 21 '18 at 3:35
user1551
71.5k566125
add a comment |
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1 Answer
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Suppose $A$ has $k$ nonzero singular values (i.e. $sigma_k>sigma_{k+1}=0$). Partition $U$ as $pmatrix{U_1&U_2}$ and $V$ as $pmatrix{V_1&V_2}$, where each of $U_1$ and $V_1$ has $k$ columns. The so-called reduced/economic SVD is given by $U_1operatorname{diag}(sigma_1,ldots,sigma_k)V_1^T$. Note that the result of this matrix product is precisely equal to $A=USV^T$. The terms "reduced" or "economic" refer not to the matrix product, but to the sizes of multiplicands: since $U_1,operatorname{diag}(sigma_1,ldots,sigma_k)$ and $V_1$ have smaller sizes than $U,Sigma$ and $V$, it is more economical to store them than to store a full SVD.
answered Nov 21 '18 at 3:35
user1551
71.5k566125
add a comment |
Suppose $A$ has $k$ nonzero singular values (i.e. $sigma_k>sigma_{k+1}=0$). Partition $U$ as $pmatrix{U_1&U_2}$ and $V$ as $pmatrix{V_1&V_2}$, where each of $U_1$ and $V_1$ has $k$ columns. The so-called reduced/economic SVD is given by $U_1operatorname{diag}(sigma_1,ldots,sigma_k)V_1^T$. Note that the result of this matrix product is precisely equal to $A=USV^T$. The terms "reduced" or "economic" refer not to the matrix product, but to the sizes of multiplicands: since $U_1,operatorname{diag}(sigma_1,ldots,sigma_k)$ and $V_1$ have smaller sizes than $U,Sigma$ and $V$, it is more economical to store them than to store a full SVD.
answered Nov 21 '18 at 3:35
user1551
71.5k566125
add a comment |
Suppose $A$ has $k$ nonzero singular values (i.e. $sigma_k>sigma_{k+1}=0$). Partition $U$ as $pmatrix{U_1&U_2}$ and $V$ as $pmatrix{V_1&V_2}$, where each of $U_1$ and $V_1$ has $k$ columns. The so-called reduced/economic SVD is given by $U_1operatorname{diag}(sigma_1,ldots,sigma_k)V_1^T$. Note that the result of this matrix product is precisely equal to $A=USV^T$. The terms "reduced" or "economic" refer not to the matrix product, but to the sizes of multiplicands: since $U_1,operatorname{diag}(sigma_1,ldots,sigma_k)$ and $V_1$ have smaller sizes than $U,Sigma$ and $V$, it is more economical to store them than to store a full SVD.
answered Nov 21 '18 at 3:35
user1551
71.5k566125
Suppose $A$ has $k$ nonzero singular values (i.e. $sigma_k>sigma_{k+1}=0$). Partition $U$ as $pmatrix{U_1&U_2}$ and $V$ as $pmatrix{V_1&V_2}$, where each of $U_1$ and $V_1$ has $k$ columns. The so-called reduced/economic SVD is given by $U_1operatorname{diag}(sigma_1,ldots,sigma_k)V_1^T$. Note that the result of this matrix product is precisely equal to $A=USV^T$. The terms "reduced" or "economic" refer not to the matrix product, but to the sizes of multiplicands: since $U_1,operatorname{diag}(sigma_1,ldots,sigma_k)$ and $V_1$ have smaller sizes than $U,Sigma$ and $V$, it is more economical to store them than to store a full SVD.
answered Nov 21 '18 at 3:35
user1551
71.5k566125
answered Nov 21 '18 at 3:35
user1551
71.5k566125
answered Nov 21 '18 at 3:35
user1551
71.5k566125
answered Nov 21 '18 at 3:35
user1551
71.5k566125
71.5k566125
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If you already know that there's a single singular value, then you also know that there's a single left singular vector and a single right singular vector, both of which are very easy to compute. Isn't that all the information you need for a reduced SVD?
– Qiaochu Yuan
Nov 21 '18 at 0:46
I thought for SVD of any given $m times n$ matrix there will be $m$ left singular vectors and $n$ right singular vectors? I've been trying to determine some kind of pattern by generating various unit matrices and using the svd function in MATLAB, but I can't discern anything reasonable.
– rcmpgrc
Nov 21 '18 at 3:14
Yes, that's true of SVD, which is non-unique. But only the singular vectors associated to the nonzero singular values matter; that's the point of the reduced SVD.
– Qiaochu Yuan
Nov 21 '18 at 4:25