## Characteristic polynomials of AB and BA

Posted: May 15, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Let $\mathbb{M}_{r,s}(\mathbb{C})$ be the set of all $r \times s$ matrices with entries from $\mathbb{C}.$ Let $I$ be the identity matrix and $f_D(x)$ the characteristic polynomial of a square matrix $D$.

Problem. Let $1 \leq m \leq n \in \mathbb{N}$ and suppose that $A \in \mathbb{M}_{m,n}(\mathbb{C})$ and $B \in \mathbb{M}_{n,m}(\mathbb{C}).$ Prove that $f_{BA}(x)=x^{n-m}f_{AB}(x).$

Solution. We’ll consider three cases:

Case 1. $n=m$ and $A$ is invertible : in this case we’ll have $BA=A^{-1}ABA.$ Thus $AB$ and $BA$ are similar and hence $f_{BA}(x)=f_{AB}(x).$

Case 2. $n=m$ and $A$ is not invertible : in this case, since the equation $\det(tI + A) = 0$ has a finitely many solutions for $t \in \mathbb{R},$ there exists some $\epsilon > 0$ such that $\det(tI+A) \neq 0,$ for all $0 < |t|< \epsilon,$ i.e. $tI + A$ is invertible for all $0 < |t| < \epsilon.$ Thus, by Case 1, for any $t$ with $0 < |t| < \epsilon$, we have

$\det(xI - (tI+A)B)=\det(xI - B(tI+A))$.

Now letting $t \to 0$ we’ll get $f_{AB}(x)=\det(xI - AB)=\det(xI - BA)=f_{BA}(x).$

Case 3. $n > m$ : in this case we add $n-m$ zero rows (resp. columns) to $A$ (resp. $B$) to get an $n \times n$ matrix $A_0$ (resp. $B_0$). It follows immediately that

$B_0A_0=BA$ and $A_0B_0=\begin{pmatrix} AB & 0 \\ 0 & 0 \end{pmatrix}$.

Therefore:

$f_{A_0B_0}(x)=\det(xI - A_0B_0)=x^{n-m} \det(xI - AB)=x^{n-m}f_{AB}(x).$

Thus,  by Case 1 and Case 2

$f_{BA}(x)=f_{B_0A_0}(x)=f_{A_0B_0}(x)=x^{n-m}f_{AB}(x). \ \Box$

Remark. So if $n=m$, then, counting multiplicity, the eigenvlaues of $AB$ and $BA$ are the same. If $n>m,$ counting multiplicity, the eigenvalues of $BA$ are the same as the eigenvalues of $AB$ plus $n-m$ eigenvalues which are all $0$.

As I mentioned in my post, $0 < |t| < \epsilon.$