# 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?

• Jan 13, 2013 at 21:39
• The identity is ${ x = 1 }$ case of this result. Nov 27, 2023 at 10:44

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.

• "simply pad-up appropriately with 0's and 1's to get square matrices." Oh, I can't believe it! Very nice! Many thanks. Jan 17, 2011 at 7:37
• There is rarely need for anything... For example, there is no need for proofs to be purely algebraic :) Jan 17, 2011 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). Jan 17, 2011 at 15:31
• "Schur decomposition" typically refers to Schur upper triangularization; it might be clearer to say that you are employing "the Schur complement". Feb 6, 2022 at 15:23
• @BruceGeorge How to extend non-square matrices by padding with 0 and 1?
– log2
Nov 10, 2023 at 10:08

(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.

• Nice argument. I guess this is very close to the "Schur decomposition" method suggested by Professor Dubuque. Jan 17, 2011 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. Jan 17, 2011 at 7:46
• @MarianoSuárez-Álvarez: Are you sure about the surjection argument? We still can't make sense of $\det A$ when $A$ is not square. Nov 19, 2018 at 8:03
• @MarianoSuárez-Álvarez I can't see the equality, could u explain? Dec 12, 2022 at 14:10

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, $$\det\begin{pmatrix} I_m & -A \\ B & I_n \end{pmatrix} = \det\begin{pmatrix} I_m+AB & 0 \\ B & I_n \end{pmatrix} = \det(I_m + AB).$$

• Maybe it's useful to show why you can do like that, that is, by multiplying on the right or on the left by the determinant $1$ matrix $\begin{pmatrix}I_m & A\\ 0 & I_n\end{pmatrix}$. Jan 5, 2017 at 11:46
• What exactly is that determinant? Its elements are matrices itself? Jan 5, 2017 at 11:50
• As I mentioned in my post in the lnked dupe, this is a special case of Schur decomposition. Jan 5, 2017 at 14:45

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.

• You need to show further that a non-zero eigenvalue has the same algebraic multiplicity for both $AB$ and $BA$. Only after that can you conclude the proof. Sep 19, 2019 at 16:02