Given a matrix $\mathbf A\in \mathbb R^{m\times n}$ and an identity matrix $\mathbf I$ of appropriate dimensions, how do you prove $\det(\mathbf I+\mathbf A^T\mathbf A) = \det (\mathbf I+\mathbf A\mathbf A^T)$?
Prove $\det(\mathbf I+\mathbf A^T\mathbf A) = \det (\mathbf I+\mathbf A\mathbf A^T)$
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1Note this is a special case of [Sylvester's deterimant theorem](http://www.ee.iitb.ac.in/spann/sylvester.pdf). – 2011-09-14
2 Answers
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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0@anon: Sorry: I should have checked. That’s a nice little exposition. – 2011-09-14
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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0I wanted to post this proof, but hesitated since the OP has [previously mentioned](http://math.stackexchange.com/questions/62213) that he doesn't know about SVD. – 2011-09-14