$A$ is complex matrix and $A^3=A$. Show that $rk(A)=tr(A^2)$

Question

$A$ is complex matrix and $A^3=A$. Show that $rk(A)=tr(A^2)$

I'm more concerned with how I can derive the prove of this question. before I ask this question, I fail to prove that whit jordan canonical form

Answer 1

If $A^3=A$ then we have

• $A^3-A = A(A^2-I)= A(A-I)(A+I)=0 \rightarrow$ the eigenvalues of $A$ can only be $0,1,-1$
• $rk(A) =rk(A^3) \leq rk(A^2) \leq rk(A) \rightarrow rk(A) = rk(A^2)$
• the eigenvalues of $A^2$ can only be $0,1$ and $tr(A^2)= rk(A^2)$ is the (algebraic) multiplicity of the eigenvalue 1. (think of the Jordan form of $A$ and $A^2$)
• $\rightarrow rk(A) = tr(A^2)$

Answer 2

Since $A^3 - A = 0$, it follows that $p(A) =0$, with $p(t) = t^3 - t = t(t+1)(t-1)$.

It is well-known that the minimal polynomial $\varphi(t)$ of $A$, which is the unique monic polynomial of minimal degree that annihilates $A$, must divide every polynomial that annihilates $A$. Thus, if $p(A) = 0$ and $\varphi$ is the minimal polynomial of $A$, then there is a polynomial $\psi$ such that $p = \varphi \psi$.

In the example above, this leaves the following possibilities: \begin{align} \varphi(t) &= t \\ \varphi(t) &= t+1 \\ \varphi(t) &= t- 1 \\ \varphi(t) &= t(t+1) \\ \varphi(t) &= t(t-1) \\ \varphi(t) &= (t+1)(t-1) \\ \varphi(t) &= t(t+1)(t-1). \end{align}

One can examine each case individually. For instance, if $\varphi(t) = t(t+1)(t-1)$, then the eigenvalues of $A$ are $\lambda_1 = 1$ with multiplicity $m_1 \ge 1$; $\lambda_2 = -1$ with multiplicity $m_2 \ge 1$; and $\lambda_3 = 1$ with multiplicity $m_3 \ge 1$. In this case, $\text{rank}(A) = m_1 + m_2$ and since the eigenvalues of $A^2$ are $\lambda_1$, with multiplicity $m_1+m_2$, and $\lambda_3$, with multiplicity $m_3$, it follows that $\text{tr}(A^2) = m_1 + m_2$ (see observations below).

Observation 1. If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then $A^nv=\lambda^n v$ for every positive integer $n.$

Observation 2. If $A$ is an $n$-by-$n$ matrix with eigenvalues $\lambda_1,\dots,\lambda_n$ (multiplicities included), then $\text{tr}(A) = \sum_{i=1}^n \lambda_1$.

Answer 3

Since $A^3-A=0$, the minimal polynomial of $A$ divides $(t-1)(t+1)t$. Hence the minimal polynomial can be factorized in distinct linear factors. This implies that all Jordan blocks of $A$ have size $1$, hence $A$ is diagonalizable.

The rank of $A$ is the number of non-zero eigenvalues of $A$, which is equal to the number of non-zero eigenvalues of $A^2$ by diagonalizability, which is equal to $tr(A^2)$ as the only non-zero eigenvalue of $A^2$ is $1$.