Suppose $A \in \Bbb R ^{n \times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $\operatorname{trace}(A^2) = \operatorname{rank}(A)$.

I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?


$Tr(A^2)=\sum_{i}\lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.

  • $\begingroup$ Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values $\endgroup$ – Learnmore Oct 6 '15 at 10:13
  • $\begingroup$ @learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues. $\endgroup$ – Stabilo Oct 6 '15 at 10:40
  • $\begingroup$ $tr(D)$ =sum of diagonal elements ;anyway got your point $\endgroup$ – Learnmore Oct 6 '15 at 10:44

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