Singular Value Decomposition of a Real Unit Matrix











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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 $n$.










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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 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 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 at 4:25















up vote
0
down vote

favorite
1












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 $n$.










share|cite|improve this question






















  • 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 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 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 at 4:25













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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 $n$.










share|cite|improve this question













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






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asked Nov 20 at 20:35









rcmpgrc

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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 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 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 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 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 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 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 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 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 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 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 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 at 4:25










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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.






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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.






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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.






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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.






        share|cite|improve this answer












        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.







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        answered Nov 21 at 3:35









        user1551

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