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If $A$ and $B$ are $n$ by $n$ matrices show that $AB$ and $BA$ have the same eigenvalues. I see why this is true if both are nonsingular. But does it still hold if they are not invertible?



marked as duplicate by mdp, Cheerful Parsnip, Git Gud, user641, Thomas Andrews May 1 '13 at 15:26

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    $\begingroup$ See this answer to see why they have the same characteristic polynomial. Algebraically. $\endgroup$ – Julien May 1 '13 at 15:17
  • $\begingroup$ @julien Copy and paste it? $\endgroup$ – Git Gud May 1 '13 at 15:18
  • $\begingroup$ @GitGud I've already copied/pasted this on the exact same question. I'm looking for the duplicate... $\endgroup$ – Julien May 1 '13 at 15:23
  • $\begingroup$ Note that the answer on the marked duplicate needs some details. What is written works to show that $AB$ and $BA$ have the same nonzero eigenvalues (as $Bv\neq 0$ necessarily in this case). So you need to treat $0$ separately. But then clearly $AB$ is invertible iff $BA$ is invertible. By det, e.g. $\endgroup$ – Julien May 1 '13 at 15:39

Fix $\lambda$. As you mentioned, if $B$ is invertible, it is easy to show that

$$\det(\lambda I-AB)=\det(\lambda I-BA) \,.$$

Now, look at the Polynomial

$$P(x)=\det[\lambda I-A(B-xI)]-\det[\lambda I-(B-xI)A] \,.$$

What is $P(x)$ when $B-xI$ is invertible? And don't forget to explain why $P$ is a polynomial.


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