Linear Algebra, matrix for a linear transformation Question 1. Let $V=\mathbb{R}^3$, $T:V \rightarrow V$ be linear. Suppose that $T^3=T, T^2 \neq T, T^2 \neq Id,$ and $\dim \ker T = 2.$ Show that the matrix of $T$ with respect to some basis is
$$
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & 0 \\
  0 & 0 & -1 \\
\end{pmatrix}.
$$
Attempt. The first two columns of $0$s come from the fact that $\dim \ker T = 2.$ Furthermore, from the hypothesis, we have
$$0 = T(T-I)(T+I)$$
where $T(T-I), (T-I)(T+I), (T-I),(T+I) \neq 0$. How to continue from here? We need to show that $-1$ is an eigenvalue. 
 A: The polynomial $P = X^3-X  = (X-1)(X+1) X$ is divisible by the minimal polynomial $\mu_T$ of $T$.
More over, from the other hypotheses, $X^2-1=(X-1)(X+1)$ and $X^2-X=(X-1) X$ are not divisible by $\mu_T$.
There is two options remaining:


*

*$\mu_T= X(X+1)(X-1)$ then $\dim(\ker(T))=1$.

*$\mu_T=X (X+1)$ so the only eigenvalues are $-1$ and $0$ and you can conclude from $\dim(\ker(T))$.

A: $$ T(T-I)(T+I) =0 \implies \lambda (\lambda-1)(\lambda+1)=0 $$
Where $\lambda$ is an eigenvalue of your matrix.
Upon the orthogonal diagonalization you will find a  diagonal matrix with the eigenvalues on the main diagonal. 
The set of eigenvectors associated with this matrix constitute the desired basis for the space. 
A: You are given that $T(T-I)(T+I)=0$, $T(T-I)\ne 0$ and $(T-I)(T+I)\ne 0$.


*

*Because $T(T-I)\ne 0$ and $(T+I)T(T-I)=0$, then $-1$ is an eigenvalue of $T$.

*Because $(T-I)(T+I)\ne 0$ and $T(T-I)(T+I)=0$, then $0$ is an eigenvalue of $T$.
The dimension of $\mbox{ker}(T)$ is $2$, which forces the dimension of $\mbox{ker}(T+I)$ to be $1$, because the space is 3-dimensional and $\mbox{ker}(T+I)\ne \{0\}$. So $T$ is diagonalizable with eignevalues $0,0,-1$.
