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I am trying to prove that similar nxn matrices A and B share the same eigenvalues and characteristic polynomial.

I am not sure how to start, but I am tempted to start with something like

$$tr(A) = tr(B)$$

and

$$det(A) = det(B)$$

since we know that the matrices are similar.

But, this approach sort of seems to be using what I am trying to prove.

Is this a valid approach? Or, should I head in another direction instead?

Thanks,

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    $\begingroup$ Is finding the characteristic polynomial part of the problem you are trying to solve or is it part of the approach you are taking to show that the eigenvalues are the same? $\endgroup$ – John Douma Nov 20 '15 at 4:59
  • $\begingroup$ the problem statement asks to show that A and B share the same eigenvalues and characteristic polynomial @JohnDouma, so I think that it is part of the problem, too ... $\endgroup$ – User001 Nov 20 '15 at 5:00
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    $\begingroup$ To show that the characteristic polynomials are the same, take a look at math.stackexchange.com/questions/87699/… $\endgroup$ – John Douma Nov 20 '15 at 5:18
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If $A$ is similar to $B$ then there exists invertible $P$ such that $A=P^{-1}BP$.

We can rewrite this as $PA=BP$.

Suppose $\lambda$ is an eigenvalue of $A$ with eignevector $v$. Then $$PAv=P(\lambda v)=\lambda Pv=BPv$$

$\therefore\lambda$ is an eigenvalue of $B$ with eigenvector $Pv$.

Switching the roles of $A$ and $B$ gives you that an eigenvalue of $B$ is an eigenvalue of $A$.

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  • $\begingroup$ Thanks so much @JohnDouma for your solution as well as the link to the pretty proof of the characteristic polynomials being the same. I got your solution after reading PA=BP, which was key - I could not have thought that myself, at least not for awhile :-). Have a great night, $\endgroup$ – User001 Nov 20 '15 at 5:42
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yet another one. Since $A$ and $B$ are similar, there exists $P$ such that $B=PAP^{-1}$. $$\det(A-\lambda I)=\det(P)\det(A-\lambda I)\det(P^{-1})= \det(B-\lambda I)$$

Thus $A$ and $B$ share same eigenvalues and characteristic polynomial.

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  • $\begingroup$ Thanks so much @dineshdileep :-) $\endgroup$ – User001 Nov 20 '15 at 5:43

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