Jordan Normal Form of $T:V\rightarrow V$ if $T^2$ is self adjoint

Question

The Question:

Suppose that $V$ is a finite dimensional inner product space over $\Bbb C$, and the linear transformation $T:V \rightarrow V$ is such that $T^2$ is self adjoint.

What are the possible Jordan Normal Forms of $T$?

---

My Attempt:

So I know that

\begin{align} T^2 \text{ is self adjoint} & \implies T^2 \text{ is diagonalisable} \\ & \implies T^2 \text{ has a basis of eigenvectors} \end{align}

And then how should I proceed from here?

Is it true that an eigenvector of $T^2$ is also an eigenvector of $T$?

Answer 1

Let $v$ be an eigenvector of $T^2$, $T^2 v=\lambda v$. Then the span $L$ of $\{v,Tv\}$ is invariant under $T$. The action of $T$ on $L$ can be represented by the matrix $$ \pmatrix{ 0 & 1\\ \lambda & 0 }. $$ Now compute the eigenvalues and vectors of this matrix. These will give you eigenvalues of $T$ with corresponding eigenvectors equal to a suitable linear combination of $v$ and $Tv$.

This shows that $T$ is diagonalizable if all eigenvalues are not zero. In addition, Jordan blocks to the zero eigenvalues are of size 2 at most.

Answer 2

No. For example, $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ is not an eigenvector of $T= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ but it is an eigenvector of $T^2\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

Remember that Jordan normal form gives a basis of generalized eigenvectors: in each Jordan block of size $m$ and eigenvalue $\lambda$, there are $m$ basis vectors $v_1, \dots, v_m$ such that $Tv_1 = \lambda v_1$, $Tv_2 = v_1 + \lambda v_2$, $Tv_3 = v_2 + \lambda v_3$, etc., i.e, $(T-\lambda)v_1=0$, $(T-\lambda)v_2=v_1$, $(T-\lambda)v_3=v_2$, etc. Assume in our case there is some block with $m \geq 3$. Then $$ \begin{align} v_1 &= (T-\lambda)v_2 \\&= (T-\lambda)^2v_3 \\&= T^2v_3 - 2\lambda Tv_3 + \lambda^2v_3 \\&= 2\lambda^2v_3 - 2\lambda Tv_3 \\&= -2\lambda(T-\lambda)v_3 \\&= -2\lambda v_2 \end{align}$$ which is impossible because the basis vectors must be linearly independent. Thus each block must have size $m \leq 2$. The example I gave above shows that size-$2$ blocks are possible, so this is as much as we can say.