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Firstly, i understand that there is a much detailed thread on this at Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!), however, they are of a much detailed concept for me to understand at my current level.

Nevertheless,

Prove that if $λ$ is an eigenvalue of $AB$ then $λ$ is also an eigenvalue of $BA$.

I wish to approach this proving question by considering $λ = 0$ and $λ ≠ 0$.

My current attempt is that if $λ = 0$, det$(AB)$ is obviously not equal to 0. And i'm basically stuck here, how does this relates to BA in having the same eigenvalue?

idolo
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    If $A$ and $B$ are not square, then $0$ may be an eigenvalue of $AB$ without it being an eigenvalue of $BA$. – Angina Seng Oct 29 '17 at 14:15
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    But yes, if $\lambda = 0$ is an eigenvalue of $AB$, then obviously, $\det(AB) = 0$. If $A$ and $B$ are square, then clearly also $\det(BA) = 0$. – amsmath Oct 29 '17 at 14:43

1 Answers1

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Hint:

$$ ABx=\lambda x \quad \Rightarrow \quad BABx=BA(Bx)=B\lambda x=\lambda (Bx) $$

Emilio Novati
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