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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?

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See also terrytao.wordpress.com/2013/01/13/… –  Andres Caicedo Jan 13 '13 at 21:39

3 Answers 3

up vote 17 down vote accepted

Hint $\ \ $ Work universally, i.e. consider the matrix entries as indeterminates $\rm\,a_{\,i\,j},b_{\,i\,j}.\,$ Adjoin them all to $\,\Bbb Z\,$ to get the polynomial ring $\rm\ R = \mathbb Z[a_{\,i\,j},b_{\,i\,j}\,].\, $ Now, in $\rm\,R,\,$ compute the determinant of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\ \ $ then cancel $\rm\ det(A)\ \ $ (which is valid since the $\,\rm R\,$ 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.

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"simply pad-up appropriately with 0's and 1's to get square matrices." Oh, I can't believe it! Very nice! Many thanks. –  Bruce George Jan 17 '11 at 7:37
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There is rarely need for anything... For example, there is no need for proofs to be purely algebraic :) –  Mariano Suárez-Alvarez Jan 17 '11 at 7:42
    
@Mariano: It's a shining example of the power of universal proofs - which deserves emphasis (esp. since this simple algebraic proof is often overlooked - even by some professional mathematicians). –  Bill Dubuque Jan 17 '11 at 15:31

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.

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(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.

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Thanks, Mariano. Density is a nice idea, but I'm afraid this argument only works when $n=m$. (I guess that's what you meant by "Start with".) –  Bruce George Jan 17 '11 at 7:31
    
in your second equation at the end there should be $AB$ instead of $AA$ –  mpiktas Jan 17 '11 at 7:40
    
Nice argument. I guess this is very close to the "Schur decomposition" method suggested by Professor Dubuque. –  Bruce George Jan 17 '11 at 7:44
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@Bruce, for the non-square situation you can argue similarly to the first part by using the fact that surjections $\mathbb R^n\to\mathbb R^m$, when $n\geq m$, are dense in the set of all matrices. –  Mariano Suárez-Alvarez Jan 17 '11 at 7:46

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