Sylvester's determinant identity

Question

Sylvester's determinant identity states that if $A$ and $B$ are matrices of sizes $m\times n$ and $n\times m$, then $$ det(I+AB)=det(I+BA), $$ where in the first case $I$ denotes the $m\times m$ identity, and in the second, the $n\times n$ identity.

Could you sketch a proof for me, or point to an accessible reference?

Answer 1

HINT $\ \ $ Work universally, i.e. over the polynomial ring $\rm\ \mathbb Z[A_{ij},B_{ij}]\ $ take the determinant of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\ \ $ then cancel $\rm\ det(A)\ \ $ (which is valid since the ring is 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, one may proceed by way of Schur decomposition, namely

$$\rm\left[ \begin{array}{ccc} 1 & \rm A \\ \rm B & 1 \end{array} \right]\ =\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm 0 & \rm 1-BA \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]$$

$$\rm\phantom{\left[ \begin{array}{ccc} 1 & \rm B \\ \rm A & 1 \end{array} \right]}\ =\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1-AB & \rm 0 \\ \rm 0 & \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]$$

See my posts in this sci.math thread on 09 Nov 2007 for further discussion.

Answer 2

(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.

Answer 3

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.