How can we prove Sylvester's determinant identity? 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?
 A: 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.
A: (1) Start, for fun, with a silly proof for square matrices:
If $A$ is invertible, then 
$$
  \det(I+AB)=\det A^{-1}\cdot\det(I+AB)\cdot\det A=\det(A^{-1}\cdot(I+AB)\cdot A)=\det(I+BA).
$$ Now, in general, both $\det(I+AB)$ and $\det(I+BA)$ are continuous functions of $A$, and equal on the dense set where $A$ is invertible, so they are everywhere equal.
(1) Now, more seriously:
$$
\det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix}
\det\begin{pmatrix}I&B\\\\0&I\end{pmatrix}
=\det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix}\begin{pmatrix}I&B\\\\0&I\end{pmatrix}
=\det\begin{pmatrix}I&0\\\\A&AB+I\end{pmatrix}
=\det(I+AB)
$$
and
$$
\det\begin{pmatrix}I&B\\\\0&I\end{pmatrix}
\det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix}
=\det\begin{pmatrix}I&B\\\\0&I\end{pmatrix}
\begin{pmatrix}I&-B\\\\A&I\end{pmatrix}
=\det\begin{pmatrix}I+BA&0\\\\A&I\end{pmatrix}
=\det(I+BA)
$$
Since the leftmost members of these two equalities are equal, we get the equality you want.
A: here is another proof of $det(1 + AB) = det(1+BA).$ We will use the fact that 
the nonzero eigen values of $AB$ and $BA$ are the same and the determinant of a matrix is product of its eigenvalues. Take an eigenvalue $\lambda \neq 0$ of $AB$ and the coresponding eigenvector $x \neq 0.$ It is claimed that $y = Bx$ is an eigenvector of $BA$ corresponding to the same eignevalue $\lambda.$
For 
$ABx = Ay = \lambda x \neq 0,$  therefore $y \neq 0.$ Now we compute $BAy = B(ABx) = B(\lambda x) = \lambda y.$  We are done with the proof.
A: We will calculate $\det\begin{pmatrix} I_m & -A \\ B & I_n \end{pmatrix}$ in two different ways. We have 
$$ \det\begin{pmatrix} I_m & -A \\ B & I_n \end{pmatrix} 
  = \det\begin{pmatrix} I_m & 0 \\ B & I_n + BA \end{pmatrix} = \det(I_n + BA). $$
On the other hand, 
$$ \det\begin{pmatrix} I_m & -A \\ B & I_n \end{pmatrix} 
  = \det\begin{pmatrix} I_m+AB & 0 \\ B & I_n  \end{pmatrix} = \det(I_m + AB). $$
