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Problem.
Suppose $A \in \mathcal{M}_{m,n}(\mathbb{C})$ and $B \in \mathcal{M}_{n,m}(\mathbb{C})$. Prove that the characteristic polynomials for the matrices $AB$ and $BA$ satisfy $\lambda^np_{AB}(\lambda) = \lambda^mp_{BA}(\lambda)$.

Hint: One way is to use Jordan forms. Alternatively, you might try to use the following block matrices: $P = \begin{pmatrix} \lambda I_m & A \\ B & I_n \end{pmatrix}$ and $Q = \begin{pmatrix} I_m & 0 \\ -B & \lambda I_n \end{pmatrix}$.

My Question.
I have no idea how to use their recommended $P$ and $Q$. Everything I've tried has led absolutely nowhere; no progress at all. I have made some progress using Jordan forms. (Specifically, that $AB$ and $BA$ share the same eigenvalues and that $\dim E_{AB, \lambda} = \dim E_{BA, \lambda}$ for each $\lambda \neq 0$.) So, same factors in $p_{AB}$ and $p_{BA}$, and same geometric multiplicities. But I can't get the final result that the algebraic multiplicities of the nonzero eigenvalues are equal. Instead, I have proved the result using a different strategy, which I have sketched below. Assuming that my proof is correct, I would still like to know how to prove this using the other methods. Any hints?

My Proof (Sketch).
I have showed that for square matrices $\bar{A}$ and $\bar{B}$, $p_{\bar{A}\bar{B}} = p_{\bar{B}\bar{A}}$. So I next assume WLOG that $A$ is wider than tall and $B$ is taller than wide and I augment $A$ and $B$ to square matrices $\bar{A}$ and $\bar{B}$ using columns or, respectively, rows of 0's. Then $\bar{A}\bar{B}$ is a block matrix with $AB$ in the upper left and 0 elsewhere, so $p_{\bar{A}\bar{B}} = p_{AB}x^{n-m}$. Then I proved that $\bar{B}\bar{A} = BA$. Thus $p_{AB}x^{n-m} = p_{\bar{A}\bar{B}} = p_{\bar{B}\bar{A}} = p_{BA}$ and the result follows.

Again, hints for the two methods suggested by the problem (or criticisms of my method) would be greatly appreciated! Thanks!

1Teaches2Learn
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    How did you show equality of the characteristic polynomials for square matrices? However, I think you should apply that for $\bar A=P$ and $\bar B=Q$ or simply something using the determinants. – Berci Mar 05 '22 at 20:55
  • @Berci Thanks for the suggestion. To answer your question: I said that for $A$ there exist invertible matrices $P$ and $Q$ such that $A = P \begin{bmatrix} I_r & 0 \ 0 & 0 \end{bmatrix} Q$. Then I let $B = Q^{-1} \begin{bmatrix} B_{11} & B_{12} \ B_{21} & B_{22} \end{bmatrix} P^{-1}$. So $AB = P \begin{bmatrix} B_{11} & B_{12} \ B_{21} & B_{22} \end{bmatrix} P^{-1}$ and $BA = Q^{-1} \begin{bmatrix} B_{11} & 0 \ B_{21} & 0 \end{bmatrix} P^{-1}$. Then $p_{AB}(\lambda) = \lambda^{n-r} p_{B_{11}}$ and $p_{BA}(\lambda) = \lambda^{n-r} p_{B_11}$. – 1Teaches2Learn Mar 05 '22 at 21:35
  • $\mathbb C$ is algebraically closed (and characteristic zero) and $\text{trace}\big((BA)^k)\big)= \text{trace}\big((AB)^k)\big)$ for all $k\in \mathbb N$ so they have the same non-zero eigenvalues -- the justification typically is Newton's Identities, but there are many other justifications – user8675309 Mar 06 '22 at 07:21
  • This question was closed as a duplicate. Although the question per se does not exactly duplicates the question in https://math.stackexchange.com/q/311342 , several answers in the linked question do address the question here. – user1551 May 03 '25 at 16:49

1 Answers1

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To directly use the hints, one may notice that: $$ \det(PQ)=\lambda^n \chi_{AB}(\lambda)$$ and $$ \det(QP)=\lambda^m \chi_{BA}(\lambda)$$

The following is another proof(related to $P,Q$), showing a natural way to get such $P,Q$.

$$ \begin{aligned} t^n \chi_{AB}(t)&= \begin{vmatrix} tI_m-AB & A \\ 0 & tI_n \end{vmatrix}\\ &= \begin{vmatrix} tI_m & A \\ tB & tI_n \end{vmatrix}\cdot \begin{vmatrix} I_m & 0 \\ -B & I_n \end{vmatrix} \qquad (\mbox{consider block-wise elementary column operations})\\ &= \det(P)\cdot \det(Q) \end{aligned}$$ And: $$ \begin{aligned} t^m \chi_{BA}(t)&= \begin{vmatrix} tI_m & A \\ 0 & tI_n-BA \end{vmatrix}\\ &= \begin{vmatrix} I_m & 0 \\ -B & I_n \end{vmatrix}\cdot \begin{vmatrix} tI_m & A \\ tB & tI_n \end{vmatrix} \qquad (\mbox{consider block-wise elementary row operations})\\ &= \det(Q)\cdot \det(P) \end{aligned}$$

Zoudelong
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