Sylvester's determinant identity states that if $A$ and $B$ are matrices of sizes $m\times n$ and $n\times m$, then
$$ \det(I_m+AB) = \det(I_n+BA)$$
where $I_m$ and $I_n$ denote the $m \times m$ and $n \times n$ identity matrices, respectively.
Could you sketch a proof for me, or point to an accessible reference?
Best Answer
Hint $ $ Work universally, i.e. consider the matrix entries as indeterminates $\:\!\rm a_{\:\!ij},b_{\:\!ij}.\,$ Adjoin them to $\,\Bbb Z\,$ to get the polynomial ring $\rm R = \mathbb Z[a_{\:\!ij},b_{\:\!ij}].\, $ In this polynomial ring $\rm R,$ compute the determinant of $\rm\, (1+A B) A = A (1+BA)\,$ then cancel the nonzero polynomial $\rm\, det(A)\, $ (valid by $\rm R$ a domain). $ $ Extend to non-square matrices by padding appropriately with $0$'s and $1$'s to get square matrices. Note that the proof is purely algebraic - it does not require any topological notions (e.g. density).
Alternatively we may employ Schur decomposition as follows
$$\rm\left[ \begin{array}{ccc} 1 & \rm A \\ \rm B & 1 \end{array} \right]\, =\, \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm 0 & \rm 1\!-\!BA \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]\qquad$$
$$\rm\phantom{\left[ \begin{array}{ccc} 1 & \rm B \\ \rm A & 1 \end{array} \right]}\, =\, \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]\ \left[ \begin{array}{ccc} \rm 1\!-\!AB & \rm 0 \\ \rm 0 & \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]\qquad$$
See this answer for more on universality of polynomial identities, universal cancellation (before evaluation) and closely relation topics, and see also this sci.math thread on 9 Nov 2007.