Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear map such that $T^3 = T$. Show that $T$ is diagonalizable (elementarily) While revising for my linear algebra exam I came across this exam problem:
Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear mapping such that $T^3 = T$ and $T^2 \neq T,T^2 \neq \text{id}_V$.
If $ \text{dim(ker(} T )) = 2$, show that the matrix of $T$ is equal to diag$(0,0,-1)$ with respect to a suitable basis.
My idea
Assume for the moment that $T$ is diagonalizable. If $\lambda$ is an eigenvalue of $T$ then $\lambda^3 = \lambda$ since $T^3 = T$, so $\lambda \in \{ 0,-1,1 \}$.
Since $ \text{dim(ker(} T )) = 2$, clearly $0$ is an eigenvalue. If all three values above would be eigenvalues then the geometric multiplicity of $0$ would be $1$ and not $2$, so $-1$ and $1$ can't both be eigenvalues.
If $1$ is the other eigenvalue then since $T$ is diagonalizable we'd have that the matrix of $T$ is $M = $ diag $(0,0,1)$ with respect to some basis. But $M^2 = M$ so we'd have $T^2 = T$ which is not the case.
So $-1$ must be the other eigenvalue and the result follows.
Question How can I prove that $T$ is diagonalizable? For some context, the only definition of diagonalizability we have is that $T: V \rightarrow V$ diagonalizable if there is a basis of $V$ made up of eigenvectors of $V$ and that's how we usually prove diagonalizability (by explicitely providing a basis). We also have the spectral theorem for symmetric matrices but I doubt it helps here.
(I also know that the minimal polynomial of $T$ must divide $x(x-1)(x+1)$ via the relation given, so the minimal polynomial is a product of distinct linear factors, so $T$ is definitely diagonalizable. But we haven't done this in the course and this method wouldn't be accepted in the exam without a full proof, so I'm looking for a more elementary way.)
 A: Very elementarily:
Choose a basis for the kernel of $T$ and extend to a basis for the entire $\mathbb R^3$. The matrix of $T$ in this basis is now
$$ \begin{pmatrix} 0 & 0 & a \\ 0 & 0 & b \\ 0 & 0 & c \end{pmatrix} $$
and then by direct computation
$$ T^2 = \begin{pmatrix} 0 & 0 & ca \\ 0 & 0 & cb \\ 0 & 0 & c^2 \end{pmatrix} 
\qquad
T^3 = \begin{pmatrix} 0 & 0 & c^2a \\ 0 & 0 & c^2b \\ 0 & 0 & c^3 \end{pmatrix} $$
Since $T^3=T$ we must have $c^3=c$, but $c\ne 1$ because $T^2\ne T$, and $c\ne 0$ because otherwise $T=T^3=0$ whose kernel is too large. Therefore $c=-1$.
This means that $-1$ is an eigenvalue, and you can replace the third basis vector by a corresponding eigenvector.
A: From $T^3=T$ we obtain $T(T+1)(T-I)v=0$ for any $v$, so
$(T^2-I)v=v_0$ is an eigenvector $v_0$ of $\lambda = 0$
$(T^2+T)v=v_1$ is an eigenvector $v_1$ of $\lambda= 1$
$(T^2-T)v=v_{-1}$ is an eigenvector $v_{-1}$ of $\lambda=-1$
(why)?
But, in that case,
$$v=\frac{1}{2}v_1 + \frac{1}{2}v_{-1} - v_0$$
Can you conclude?
A: Let $(v_1, v_2)$ be a basis of $\ker T$ and let $\text{im }T = \text{span } v_3  \ne 0$
It is clear that $(v_1, v_2, v_3)$ is an eigenbasis...
For otherwise $v_3 \in \ker T$ so $T^2 = 0$. Thus $T = T^3 = 0$,  contradiction.
