Prove that $r(A)=\operatorname{tr}(A^2)$

Question

Let $A\in M_n(\mathbb{C})$. Show that if $A^3=A$, then $r(A)=\operatorname{tr}(A^2)$.

Since $A^3=A$, the possible eigenvalues are $0,1,-1$. I don't know from here how to compute the rank of $A$.

Edited

Since eigenvalues of $A$ are $0,1,-1$ SO eigenvalues of $A^2=0,1$ So $r(A^2)=tr(A^2)$ Now we have to show $r(A^2)=r(A)$ where rank of $A^2$ is the number of non-zero eigenvalues .

See also this question. The arguments are the same. – Dietrich Burde Dec 07 '18 at 19:39 — Dietrich Burde, Dec 07 '18 at 19:39

score 3 · Answer 1 · answered Dec 07 '18 at 19:29

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The rank is the number of nonzero eigenvalues and as their squares are 0 or 1, this number is just the same as the sum of the squares.

answered Dec 07 '18 at 19:29

Richard Martin

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but why both equal ?? – RAM_3R Dec 07 '18 at 19:34
@RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank. – Zach Langley Dec 07 '18 at 19:38
@RAM_3R Because the trace is the sum of the eigenvalues – Alan Dec 07 '18 at 19:39
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This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable. – user1551 Dec 07 '18 at 19:44
It's the number of non-zero eigenvalues counted with multiplicity. – Bernard Dec 07 '18 at 19:46
@Zach Langley then $r(A^2)=tr(A^2)$ – RAM_3R Dec 07 '18 at 19:47
why $r(A)=tr(A^2)$ – RAM_3R Dec 07 '18 at 19:50

Prove that $r(A)=\operatorname{tr}(A^2)$

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