Proving that if $A$ is a $8\times 8$ matrix over $\mathbb{R}$ and $A^3=A$, then $A$ is diagonalizable.

Question

If $A$ is a $8\times 8$ matrix over $\mathbb{R}$ and $A^3=A$ then prove that $A$ is diagonalizable.

I have got that the minimal polynomial of $A$ may be $x^3-x$ or $x$ or $x(x+1)$ or $x(x-1)$ in the 2nd case it is not possible in the 3rd and 4th cases the matrix will be the identity matrix or identity matrix multiplied by a scalar!

But I cannot reach any further.

Answer 1

For any $v$, since $A(A-I)(A+I)v=0$, you get that:

1. $v_{1}=(A^2+A)v$ is either zero or an eigenvector for eigenvalue $1$.
2. $v_{-1}=(A^2-A)v$ is either zero or an eigenvector for eigenvalue $-1$.
3. $v_0=(A^2-I)v$ is either zero or an eigenvector for eigenvalue $0$.

Now, note that:

$$v=\frac{1}{2}v_1+\frac{1}{2}v_{-1}-v_0$$

Why does this show that $A$ is diagonalizable?

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In general, if $p(A)=0$ for some $p(x)=(x-a_1)(x-a_2)\cdots (x-a_k)$ for distinct $a_i$ (thus, not repeat roots) then $A$ can be diagonalized.

Specifically, if we define $p_i(x)=\prod_{j\neq i} (x-a_j)$ then we get that:

$$1=\sum_{i=1}^{n} \frac{1}{p_i(a_i)}p_i(x)$$

So if $v_i=p_i(A)v$ then $v_i$ is an eigenvector for eigenvalue $a_i$ and:

$$v = \sum \frac{1}{p_i(a_i)} v_i$$