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I would like to show the following problem from linear algebra:

Let $n\in\mathbb{N}$ and let $A,B\in M_n$ be Hermitian matrices. If all eigenvalues of $A$ and $B$ are positive and $A^k=B^k$ for some $k\in\mathbb{N}$, then $A=B$.

I could understand that if $C$ is a Hermitian matrix, then eigenvalues of $C$ are real, and if $k=1$, then the statement is clear.

But I could not go further. I appreciate any advice. Thank you in advance.

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  • $\begingroup$ Hint. You can define the $k$-th root of any positive semi-definite Hermitian matrices. $\endgroup$ – Sangchul Lee Jul 29 '17 at 10:59
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We have

$$\ker(A-\lambda I_n)\subset \ker (A^k-\lambda^k I_n)$$ and since $A$ is Hermitian then diagonalizable so $$n=\dim\Bbb C^n=\sum_{\lambda\in\operatorname{Sp}(A)}\dim\ker(A-\lambda I_n)$$ Moreover, since $\lambda^k$ are distinct then $\sum\ker(A^k-\lambda^k I_n$) is a direct sum and so from the first inclusion we see that in fact $$\ker(A-\lambda I_n)=\ker(A^k-\lambda^k I_n)$$ The same reasoning for $B$ gives us finally $$\ker(A-\lambda I_n)=\ker(B-\lambda I_n)$$ and since $A$ and $B$ are diagonalizable we get $A=B$.

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    $\begingroup$ Thank you for the answer! But where did you use the hypothesis that eigenvalues are positive? $\endgroup$ – Tom TJ Jul 29 '17 at 11:47
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    $\begingroup$ The eigenvalues are positive then $\lambda^k$ are distinct whatever the value of $k$. $\endgroup$ – user296113 Jul 29 '17 at 13:28

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