# 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.
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.