Suppose that one has two matrices $A$, $B$. Then

Prove that $$|AB - \lambda I| = |BA - \lambda I|,$$ where $|\cdot|$ denotes the determinant, $I$ - identity matrix and $\lambda \in \mathbb{C}$.

Note that $A$ and $B$ are not necessary invertible. For invertible matrices I easily found $$|AB - \lambda I| = |B(AB - \lambda I)B^{-1}| = |BA - \lambda I|.$$

  • $\begingroup$ Note that the invertible matrices are dense in the set of matrices (for the topology induced by a norm). $\endgroup$ – Marko Karbevski Feb 22 '17 at 13:13

I'm aware of three proofs. The first is the one you get by combining your argument with Marko's comment. The second is to show that $AB$ and $BA$ have the same non-zero eigenvalues with the same multiplicities. The third proof runs as follows. Set \[ M=\begin{pmatrix}I&A\\ B&I\end{pmatrix},\quad N=\begin{pmatrix}I&0\\-B&I\end{pmatrix} \] and then note that \[ MN=\begin{pmatrix}I-AB&A\\0&I\end{pmatrix},\quad NM=\begin{pmatrix}I&A\\0&I-BA\end{pmatrix}. \] Since $\det(MN)=\det(NM)$, the result follows.

One advantage of the third proof is that it works for matrices over a commutative ring.

  • $\begingroup$ @lhf Because $\det$ preserves mutliplication: $\det(MN)=\det(M)\det(N)=\det(N)\det(M)=\det(NM)$. $\endgroup$ – freakish Feb 22 '17 at 13:41
  • $\begingroup$ Could you please explain how $|I-AB|=|I-BA|$ is equivalent to the result? $\endgroup$ – Itay4 Feb 22 '17 at 13:51
  • $\begingroup$ @Itay: Replace $A$ by $\lambda^{-1}A$. (Note that $A$ And $B$ need not be square.) $\endgroup$ – Chris Godsil Feb 22 '17 at 14:03

There is an old-fashioned proof depending only on the properties of determinants and minors.

(i) In $\det(xI-X)$ the coefficient of $x^{n-k}$ is the sum of the principal $k\times k$ minors of $X$.

(ii) Let $X^{(k)}$ denote the matrix of $k\times k$ minors of $X$ for any square $X$. Then by a theorem sometimes called the Binet-Gauss Theorem we have that $(XY)^{(k)}=X^{(k)}Y^{(k)}$.

(iii) For any square matrices $X,Y$ we have $\text{tr}(XY)=\text{tr}(YX)$.

So the coefficient of $x^{n-k}$ in $\det(xI-AB)$ is $$\text{tr}((AB)^{(k)})=\text{tr}(A^{(k)}B^{(k)})$$ whereas the coefficient of $x^{n-k}$ in $\det(xI-BA)$ is $$\text{tr}((BA)^{(k)})=\text{tr}(B^{(k)}A^{(k)})$$ and these are equal.


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