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My book offers a different proof for the following question and after some consideration I reckon my attempt is incorrect, but I cannot pinpoint why.

Let A,B be square matrices of the same size. Show that the eigenvalues of AB and BA are equal.

I wrote $det(AB - \lambda I) = det(AB) - det(\lambda I) = det(BA) - det(\lambda I) = det(BA - \lambda I)$ ergo the characteristic polynomials are equal.

I ask: flaw in reasoning is where? Thanks.

Krpcannon
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2 Answers2

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The flaw is while applying determinants,see that $det(A+B)\neq det(A) + det(B)$ in general. Hope this helps!

BAYMAX
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  • Thought there was linearity. Woops. – Krpcannon Feb 06 '17 at 13:43
  • @Krpcannon the following is true , $det(AB) = det(A).det(B)$ , and other is $Trace(A+B) = Trace(A) + Trace(B)$ , where $A$ and $B$ are two matrices of same size.and also i think you would have figured out that $det(AB) = det(BA)$ – BAYMAX Feb 06 '17 at 13:46
  • Also you can see some other properties of determinants here -http://www.math.uiuc.edu/~llpku/2007fall415/hw912.pdf – BAYMAX Feb 06 '17 at 13:55
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In general, $\det(X+Y)\ne\det(X)+\det(Y)$, but in your case it could be true.

However, if the identity you use were true, you'd get $$ \det(AB-\lambda I)=\det(AB)-\det(\lambda I)=\det(AB)-\lambda^n $$ which is clearly absurd: take $B=I$ and $A$ any matrix with eigenvalues $1$ and $2$. (The identity is valid for $n=1$, actually, but it's not so useful in that case.)

egreg
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