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Let $A$ and $B$ are matrices on field $F$ of size $m \times n$ and $n \times m$ respectively. Can we claim, that nonzero eigenvalues of matrices AB and BA are the same and each of them has the same multiplicity for this matrices if:

a) $F = \mathbb{C}$

b) $F$ — arbitrary field?

minority169
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1 Answers1

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If I understand you correctly, here's a proof that every eigenvalue of AB is an eigenvalue of BA: (only if m=n)

Let $A,B$ be 2 matrix. Then, if $\lambda = 0$ :

$|AB| = 0 \rightarrow |A||B| = 0 \rightarrow |B||A| = 0 \rightarrow |BA|=0 \rightarrow 0$ is eigenvalue of BA

Otherwise, $\lambda \neq 0$:

Let v be eigenvector of AB that matches $\lambda$, then: $B(AB)v = B\lambda v = \lambda Bv$

Let: $w=Bv$, so $w \neq 0 $ because:

$A(Bv)=\lambda v=Aw=A0 \rightarrow \lambda v=0$, but $\lambda\neq0$ and $v\neq0$, then $w\neq 0$

Finally, we get:

$BA(Bv)=BAw=\lambda Bv=\lambda w \rightarrow \lambda $ is eigenvalue of $BA$

ekosman
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