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 $$A = U\Sigma V^T$$? 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 $$n$$.

• 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

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.