Consider the linear transformation $T: \mathbb{R}^{2\times 2} \rightarrow \mathbb{R}^{2\times 2}$ such that $T(A)=A-A^T$ 
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*Find a basis for $\ker T$.

*Find a basis for $\operatorname{im} T$.

*Prove that $T$ is diagonalizable. Find an eigenbasis $\mathcal{B}$ for $T$ and find the $\mathcal{B}$-matrix $[T]_\mathcal{B}$.
My attempt:
For 1:
If $A = \begin{pmatrix} a&b\\c&d \end{pmatrix}$, then $T(A) = \begin{pmatrix} a&b\\c&d \end{pmatrix}-\begin{pmatrix} a&c\\b&d \end{pmatrix} = \begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}$. Setting this equal to $0$, we have that $b = c$. Since $\mathbb{R}^{2\times 2}$ is isomorphic to $\mathbb{R}^4$, we can write $\begin{pmatrix} a\\b\\b\\d \end{pmatrix}$ as the kernel, and find a basis: $\begin{pmatrix} a\\b\\b\\d \end{pmatrix} = a\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}+b\begin{pmatrix} 0\\1\\1\\0 \end{pmatrix}+c\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix}$. So a basis for $\ker T = \{ \begin{pmatrix} 1&0\\0&0 \end{pmatrix}, \begin{pmatrix} 0&1\\1&0 \end{pmatrix}, \begin{pmatrix} 0&0\\0&1 \end{pmatrix} \}$.
For 2:
We found that the $\operatorname{im} A$ has matrices of the form $\begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}$, so that the diagonals are negatives of each other. A basis for $\operatorname{im} A$ would thus be $\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}$.
For 3:
Now I don't know how we can prove that $T$ is diagonalizable. I know we have to find an eigenbasis for it to be diagonalizable, but I don't know how to find the eigenvalues.

You already know three eigenvectors. Since your space is four-dimensional, you just need to find a fourth. Here's a hint: if $Tv = \lambda v$ and $\lambda \neq 0$, this means that $v$ must be in $T$'s....
