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In the book Linear Algebra Done Right (Axler, 237), the following is the statement of the theorem that gives us the existence of a singular value decomposition:


7.51 $\qquad$ Singular Value Decomposition

Suppose $T\in\mathcal{L}(V)$ has singular values $s_{1},\ldots,s_{n}$. Then there exist orthonormal bases $e_{1},\ldots,e_{n}$ and $f_{1},\ldots,f_{n}$ of $V$ such that

$$Tv=s_{1}\langle v,e_{1} \rangle f_{1}+\cdots+s_{n}\langle v,e_{n} \rangle f_{n}$$

for every $v\in V$.


Now, I have seen elsewhere that we can do this for general rectangular matrices, yet this statement of the theorem seems to indicate that it can only be done for square matrices (since it says $T\in \mathcal{L}(V)$ rather than $T\in \mathcal{L}(V,W)$.

Is it enough to know that you can do it to a square matrix to know that you can do it to a rectangular matrix, since you can add enough rows/columns of all $0$'s to make a rectangular matrix square? If so, why does doing this not affect the decomposition? I feel like I'm missing something obvious, but I appreciate any insight on the matter.

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    $\begingroup$ I stated the Singular Value Decomposition just for linear maps from a vector space to itself because the result appears in a chapter that focuses on operators mapping a vector space into itself, and I wanted to keep things simple. The proof given in the book easily extends to the more general context that you mention because if T maps V to W, then T^* T maps V to V. Last year I recorded a note to myself with a reminder to mention this more general context in the next edition of the book. $\endgroup$ – Sheldon Axler Mar 29 '18 at 7:36
  • $\begingroup$ Thank you! By the way, your book is great! $\endgroup$ – dirtydivider Mar 29 '18 at 13:21

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