Prove $\det(\mathbf I+\mathbf A^T\mathbf A) = \det (\mathbf I+\mathbf A\mathbf A^T)$

Given a matrix $\mathbf A\in \mathbb R^{m\times n}$ and an identity matrix $\mathbf I$ of appropriate dimensions, how do you prove $\det(\mathbf I+\mathbf A^T\mathbf A) = \det (\mathbf I+\mathbf A\mathbf A^T)$?

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Note this is a special case of Sylvester's deterimant theorem. –  anon Sep 14 '11 at 6:07

Consider the product $$\left(\begin{matrix}I_m&0_{m\times n}\\A^T&I_n\end{matrix}\right)\left(\begin{matrix}I_m+AA^T&A\\0_{n\times m}&I_n\end{matrix}\right)\left(\begin{matrix}I_m&0_{m\times n}\\-A^T&I_n\end{matrix}\right);$$ use the facts that $\det(AB)=\det(A)\det(B)$ and that the determinant of a block-triangular matrix is the product of the determinants of the diagonal blocks.

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Nice! How did you come up with this? (To be precise, you're using the fact that the determinant of a block-triangular matrix is the product of the determinants of the diagonal blocks.) –  joriki Sep 14 '11 at 6:12
@joriki: I cheated. :-) As anon just pointed out, it’s a special case of Sylvester’s determinant theorem, which is mentioned in the WP article on determinants. The WP article that he cites mentions that it’s closely related to the matrix determinant lemma, which that article proves in a special case. I just checked that the proof generalized in the obvious way. –  Brian M. Scott Sep 14 '11 at 6:19
@Brian: Your assumption that I linked to Wikipedia turns out to be false. :) (But I knew your answer looked very familiar as I had glanced the WP article on that lemma not a minute before you posted.) –  anon Sep 14 '11 at 6:25
@anon: Sorry: I should have checked. That’s a nice little exposition. –  Brian M. Scott Sep 14 '11 at 6:32

This follows from the relationship between the singular values of $A$ and the eigenvalues of $AA^T$ and $A^TA$. The latter are the same except the larger of the two has extra zeros. Since they're both symmetric, you can diagonalize them (which of course keeps $I$ diagonal), and then the given determinants are $\prod_i(1+\lambda_i)$, and the extra zero eigenvalues in the larger matrix don't make a difference.

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I wanted to post this proof, but hesitated since the OP has previously mentioned that he doesn't know about SVD. –  Ｊ. Ｍ. Sep 14 '11 at 6:20