Suppose that a map $T$ from the space of $3 \times 3$ complex matrices into itself is defined by $T(M)=M-M^*$ for $M\in\mathbb{C}^{3\times3}$. Here $M^*$ is the Hermitian transpose of $M$. What is the Jordan Canonical Form of T?
Ideas and observations: So we want to find the eigenvalues, algebraic multiplicities for how many times they show up, and then the geometric multiplicities for the size of the Jordan blocks.
For visualization: \begin{align*} M = \begin{pmatrix} a_{11} + b_{11}i & a_{12} + b_{12}i & a_{13} + b_{13}i\\ a_{21} + b_{21}i & a_{22} + b_{22}i & a_{23} + b_{23}i\\ a_{31} + b_{31}i & a_{32} + b_{32}i & a_{33} + b_{33}i \end{pmatrix}\,,\, M^* = \begin{pmatrix} a_{11} - b_{11}i & a_{21} - b_{21}i & a_{31} - b_{31}i\\ a_{12} - b_{12}i & a_{22} - b_{22}i & a_{32} - b_{32}i\\ a_{13} - b_{13}i & a_{23} - b_{23}i & a_{33} - b_{33}i \end{pmatrix} \end{align*} Then applying $T$ once should give: \begin{align*} \begin{pmatrix}2ib_{11}&\left(a_{12}-a_{21}\right)+i\left(b_{21}+b_{12}\right)&\left(a_{13}-a_{31}\right)+i\left(b_{31}+b_{13}\right)\\ \left(a_{21}-a_{12}\right)+i\left(b_{21}+b_{12}\right)&2ib_{22}&\left(a_{23}-a_{32}\right)+i\left(b_{23}+b_{32}\right)\\ \left(a_{31}-a_{13}\right)+i\left(b_{31}+b_{13}\right)&\left(a_{32}-a_{23}\right)+i\left(b_{23}+b_{32}\right)&2ib_{33}\end{pmatrix} \end{align*} And applying it once more: \begin{align*} \begin{pmatrix}4ib_{11}&\left(2a_{12}-2a_{21}\right)+i\left(2b_{21}+2b_{12}\right)&\left(2a_{13}-2a_{31}\right)+i\left(2b_{31}+2b_{13}\right)\\ \left(-2a_{12}+2a_{21}\right)+i\left(2b_{21}+2b_{12}\right)&4ib_{22}&\left(-2a_{32}+2a_{23}\right)+i\left(2b_{23}+2b_{32}\right)\\ \left(-2a_{13}+2a_{31}\right)+i\left(2b_{31}+2b_{13}\right)&\left(2a_{32}-2a_{23}\right)+i\left(2b_{23}+2b_{32}\right)&4ib_{33}\end{pmatrix} \end{align*} One observation here is that $T(T(M))$ = $2T(M)$.
Now, back to just $T(M)$. Let's split up the real and imaginary part: \begin{align*} T(M) &= \begin{pmatrix}2ib_{11}&\left(a_{12}-a_{21}\right)+i\left(b_{21}+b_{12}\right)&\left(a_{13}-a_{31}\right)+i\left(b_{31}+b_{13}\right)\\ \left(a_{21}-a_{12}\right)+i\left(b_{21}+b_{12}\right)&2ib_{22}&\left(a_{23}-a_{32}\right)+i\left(b_{23}+b_{32}\right)\\ \left(a_{31}-a_{13}\right)+i\left(b_{31}+b_{13}\right)&\left(a_{32}-a_{23}\right)+i\left(b_{23}+b_{32}\right)&2ib_{33}\end{pmatrix}\\ &=\begin{pmatrix}0 & a_{12}-a_{21} & a_{13}-a_{31}\\ a_{21}-a_{12} & 0 & a_{23}-a_{32}\\ a_{31}-a_{13} & a_{32}-a_{23} & 0 \end{pmatrix} + \begin{pmatrix}2ib_{11} & i(b_{21}+b_{12}) & i(b_{31}+b_{13})\\ i(b_{21}+b_{12}) & 2ib_{22} & (b_{23}+b_{32})\\ (b_{31}+b_{13}) & (b_{23}+b_{32}) & 2ib_{33} \end{pmatrix} \end{align*} Another observation: the real part is skew symmetric and the complex part is symmetric. Any ideas where to go from here? Thank you.
Your observation that $T^2-2T=0$ is very good. Indeed, this implies that $T$ is diagonalizable since its minimal polynomial splits into distinct linear terms.
More generally, note that the space of $n\times n$ matrices (as a real vector space of dimension $2n^2$) is given by a direct sum over Hermitian and anti-Hermitian matrices. It should be clear that $T$ takes any Hermitian matrix to zero and satisfies $T(M)=2M$ for any anti-Hermitian $M$. So $T$ is diagonalizable with eigenvalue zero appearing with multiplicity $n^2$ (the dimension of the subspace of Hermitian matrices) and eigenvalue $2$ appearing with multiplicity $n^2$ (the dimension of the subspace of anti-Hermitian matrices).