Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
See also… – Andrés E. Caicedo Jan 13 '13 at 21:39
up vote 18 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.

share|cite|improve this answer
"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
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

(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:

$$ \det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix} \det\begin{pmatrix}I&B\\\\0&I\end{pmatrix} =\det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix}\begin{pmatrix}I&B\\\\0&I\end{pmatrix} =\det\begin{pmatrix}I&0\\\\A&AB+I\end{pmatrix} =\det(I+AB) $$


$$ \det\begin{pmatrix}I&B\\\\0&I\end{pmatrix} \det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix} =\det\begin{pmatrix}I&B\\\\0&I\end{pmatrix} \begin{pmatrix}I&-B\\\\A&I\end{pmatrix} =\det\begin{pmatrix}I+BA&0\\\\A&I\end{pmatrix} =\det(I+BA) $$

Since the leftmost members of these two equalities are equal, we get the equality you want.

share|cite|improve this answer
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
@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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.