Suppose that $A$ is a $n \times n$ matrix with $A^3 = A$. Then prove that $A$ is diagonalizable.
I have proved it using the concept that minimal polynomial of $A$ consists of only linear factors. But without applying this method I couldn't able to solve it in an alternative way.
Please let me aware if there were any alternative way of solving this problem.
Thank you in advance.
Here's an interesting approach: note that $$ x = (I - A^2)x + \frac 12(A^2 - A)x + \frac 12(A^2 + A)x $$ Because $A^3 - A = 0$, note that $(I - A^2)x \in \ker(A)$, $(A^2 - A)x \in \ker(A + I)$, and $(A^2 + A)x \in \ker(A - I)$. Thus, we have shown that any vector in $\Bbb C^n$ can be decomposed into a linear combination of eigenvectors.