Jordan canonical form linear operators $T$ satisfying (separately) $T^2=T$, $T^2=1$, and $T^2=0$

Question

Let $T:V\rightarrow V$ be a a linear operator; then I need to find Jordan canonical form of each of the following cases:

1. $T^2=T$.
2. $T^2=1$;
3. $T^2=0$.

For the first and second case I know the eigenvalue are distinct so they are diagonalizable, but the eigenvalues for third one are not distinct so I am not sure. How this information will help us finding the Jordan canonical form of each operator?

Thanks for the help.

Answer 1

Hint For case (3), an eigenvector $\bf v$ of $T$, say, of eigenvalue $\lambda$ satisfies $$0 = T^2 {\bf v} = T(T {\bf v}) = T(\lambda {\bf v}) = \lambda T {\bf v} = \lambda (\lambda {\bf v}) = \lambda^2 {\bf v},$$ so $\lambda = 0$. So, the only eigenvalue of $T$ is zero.

Thus, the Jordan canonical form of $T$ is a direct sum $\bigoplus J_{k_a}(0)$ of Jordan blocks of eigenvalue zero. Since $T^2 = 0$, we must have $0 = \left(\bigoplus J_{k_a}(0)\right)^2 = \bigoplus \left(J_{k_a}(0)^2\right)$ and so each Jordan block $J_{k_a}(0)$ must satisfy $$J_{k_a}(0)^2 = 0.$$ But this is only true for Jordan blocks of size $k_a \leq 2$, so the Jordan canonical form of $T$ is a direct sum of blocks $$J_1(0) = \pmatrix{0} \qquad \textrm{and} \qquad J_2(0) = \pmatrix{0&1\\0&0} .$$ On the other hand, any such direct sum has square $0$, so these direct sums are precisely the Jordan canonical forms of linear transformations $T$ satisfying $T^2 = 0$.