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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 $A = U\Sigma V^T$? 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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  • $\begingroup$ 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? $\endgroup$ – Qiaochu Yuan Nov 21 '18 at 0:46
  • $\begingroup$ 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. $\endgroup$ – rcmpgrc Nov 21 '18 at 3:14
  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Nov 21 '18 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 $U_1\operatorname{diag}(\sigma_1,\ldots,\sigma_k)V_1^T$. 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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