$A^3 = I$ and $A ≠ I$, show A not diagonalizable

Question

I was able to show that $A^3 - I = 0$ then $(A-I)(A^2+A+I) = 0 $ then $A^2+A+I =0$, but as to show why it means A is (or isn't) diagonalizable over $R$, I'm stuck.

any help in directions for the proof will be appreciated!

Answer 1

Here is a proof that if $A$ is a matrix with real entries satisfying $A^3 = I$ with $A \neq I$, then $A$ is not diagonalizable over $\Bbb R$.

First, we note that $A^2 + A + I$ is not invertible. Indeed, suppose that this matrix is invertible. It would follow that $$ A^3 - I = 0 \implies (A^2 +A + I)(A - I) = 0 \implies A - I = 0 \implies A = I, $$ contradicting our assumption that $A \neq I$.

Now, let $\omega_1,\omega_2$ denote the two (non-real) complex roots of $\omega^2 + \omega + 1 = 0$. We have $$ \det(A^2 + A + I) = 0 \implies \det((A - \omega_1 I)(A - \omega_2 I)) = 0 \\ \implies \det(A - \omega_1 I)\det(A - \omega_2 I) = 0. $$ Thus, it must holds that $\det(A - \omega_1 I) = 0$ or $\det(A - \omega_2 I) = 0$. Thus, $A$ has an eigenvalue that is not real. Thus, $A$ cannot be diagonalizable over $\Bbb R$.

Answer 2

As a counterexample take the matrix describing a rotation of $120°$. For dimension higher than two fill the rest of the matrix with $1$'s in the diagonal and zero elsewhere.