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If $A$ is diagonalizable then $\mbox{rank}(A) = \mbox{tr}(A^2)$.

How to find a counter example for a matrix $A^3=A$ but $\mbox{rank}(A) \neq \mbox{tr}(A^2)$

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    $\begingroup$ What is the field over which you are working? Over the finite field with two elements, the identity matrix of any size $n \geq 2$ has trace $0$ or $1$, but its rank is $n$. $\endgroup$ Jun 15 '20 at 12:20
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There is no counterexample over a field of characteristic $0$, i.e., fields which contain $\mathbb Q$. Proof:

Suppose that $A^3 = A$. This means that $A$ is a root of the polynomial $p(t) = t(t-1)(t+1)$, hence the minimal polynomial of $A$ divides $p(t)$. Since this minimal polynomial splits as a product of distinct linear factors (since $1+1 \neq 0$), we obtain that $A$ is diagonalizable.

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Over a field $\Bbb{F}$ of characteristic $\neq2$ such a matrix is automatically diagonalizable. Consequently you cannot find counterexamples.

A quick way to see this is that any vector $x\in\Bbb{F}^n$ ($n$ = the size of the matrix $A$) can be written as a linear combination of eigenvectors: $$ x=(x-A^2x)+\frac12(A^2x+Ax)+\frac12(A^2x-Ax). $$ The vectors in parens are easily seen to be eigenvectors of $A$ belonging to eigenvalues $\lambda_1=0$, $\lambda_2=+1$ and $\lambda_3=-1$ respectively.

When the eigenvectors of a matrix span the entire space, the matrix is diagonalizable.

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  • $\begingroup$ Cool argument. Not OP, but may I ask how you found the eigenvector summands? $\endgroup$ Jun 15 '20 at 12:46
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    $\begingroup$ @RichardJensen A bit of experience is enough with such a simple minimal polynomial :-) Actually I think there is general formula for this. Here the polynomial is $p(T)=T^3-T$. Its maximal factors are $p_1(T)=T^2-T$, $p_2(T)=T^2-1$ and $p_3(T)=T^2+T$. And you see them appearing. We can write $1$ as their linear combination. Finding the coefficients is actually equivalent to decomposing the reciprocal as a partial fraction: $$\frac1{x^3-x}=\frac A x+\frac B{x-1}+\frac C{x+1}.$$ This is possible whenever the roots are distinct. $\endgroup$ Jun 15 '20 at 12:53
  • $\begingroup$ I see, thanks you very much for your answer! $\endgroup$ Jun 15 '20 at 15:58
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There is not any counterexample, regardless of the underlying field or whether $A$ is diagonalisable. When $A^3=A$, $\operatorname{rank}(A)$ must be equal to $\operatorname{tr}(A^2)$. (When the field has characteristic $p>0$, this identity should be understood as one over $\mathbb F$. That is, we actually mean $\varphi(\operatorname{rank}(A))=\operatorname{tr}(A^2)$ where $\varphi$ is the ring homomorphism between $\mathbb Z$ and $\mathbb F$.)

Denote the underlying field by $\mathbb F$ and let $A$ be $n\times n$. Since $(x^2-1)x$ is an annihilating polynomial of $A$ and $x^2-1,x$ are relatively prime, $\mathbb F^n = \ker((A^2-I)A) = \ker(A^2-I)\oplus\ker(A)$. Also, as $A^2-I$ and $A$ commute with $A$, both $\ker(A^2-I)$ and $\ker(A)$ are invariant subspaces of $A$. Therefore \begin{aligned} \operatorname{tr}(A^2) &= \operatorname{tr}(A^2|_{\ker(A^2-I)}) + \operatorname{tr}(A^2|_{\ker(A)})\\ &= \operatorname{tr}(A^2|_{\ker(A^2-I)})\\ &= \operatorname{tr}(I|_{\ker(A^2-I)}) \quad(\because A^2=I \text{ on } \ker(A^2-I))\\ &= \dim\ker(A^2-I) \quad\text{(we mean $\varphi(\dim\ker(\cdots))=\operatorname{tr}(\cdots)$ here)}\\ &= n - \dim\ker(A) \quad(\because \mathbb F^n=\ker(A^2-I)\oplus\ker(A))\\ &= \operatorname{rank}(A). \end{aligned}

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    $\begingroup$ I couldn't find a counterexample in characteristic two. Now I know the reason! $\endgroup$ Jun 15 '20 at 15:32

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