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If I take a common eigen vector, I come to the conclusion that $a(b-\frac{1}{2})=0$ with (a,b) 2 eigen values. a for A and b for B which invited me first to see Ker(A).

If $0$ is an eigen value for A, the problem is done because B(Ker(A)) $\subset$ Ker(A) and B belong to Mn(C) so B restricted to Ker(A) has an eigen value and then a common eigen vector... (characteristic polnomial has a root, because $dim(Ker(A))>0$

But if $0$ isn't an eigen value of $A$, I do not know what to do and how to use the oddness of $n$. $A^2$ and $B$ commute. If $0$ isn't an eigen vector for A, then $I=B+A^{-1}BA$ but it doesn't seem to help a lot... If $n$ is odd then A and B have at least real value (for B it could be that $b=\frac{1}{2}$...) and I find with my egality that $tr(B)=\frac{n}{2}$

What do you recommend me to do?

Dlouna.J
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  • For more context, see also here. – Dietrich Burde Nov 11 '23 at 11:26
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    You can also find a 'guide post' solution here.https://math.stackexchange.com/questions/2786438/ab-ba-a-implies-a-and-b-have-a-common-eigenvector – brighton Nov 11 '23 at 11:32
  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Nov 11 '23 at 11:35
  • Yes thanks. The idea is to use this $\frac{1}{2}$ as a hint and make it appear and then we're done since $A(Ker(B-\frac{1}{2}I)) \subset Ker(B-\frac{1}{2}I)$ – Dlouna.J Nov 11 '23 at 12:58
  • Notice that if $A$ is invertible then $B$ and $I-B$ are similar, so they have the same characteristic polynomial, on the other hand, can you relate the characteristic polynomials of $B$ and $I-B$? – Mor A. Nov 11 '23 at 14:40
  • $\chi_{I-B} (X) =det(XI-(I-B)=det( (X-1)I+B)=-det((1-X)I-B)=-\chi{B}(1-X)$ because $n$ is odd... So $\chi_B (X) + \chi_B (1-X)=0$ and $\frac{1}{2}$ is a root

    Taking an eigen vector $x_{\frac{1}{2}}$, $ \frac{1}{2}AX=BAX$ and its done. Was it what you were waiting for? I'm interested to know what you were thinking about...

     – Dlouna.J Nov 11 '23 at 16:02

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