Basis for $\ker(T)$, where $T(A)=A^T-A$

Question

I just finished my linear algebra final and one question has been bugging me. We were asked to find the basis for $\ker(T)$, where $T$ is the linear transformation $T(A)=A^T-A$.

It went on to ask a few other things, dimension, rank etc.

here's what I did:

If $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$, then $$A^T-A=\begin{pmatrix} 0 & c-b \\ b-c & 0\end{pmatrix}$$ this resulted in $(b-c)x_1=0$ and $(c-b)x_2=0$.

I feel like I went wrong with this problem and I cant find anything similar too it?

Answer 1

It follows from your computations that $\ker T$ is the space of the $2\times2$ symmetric matrices. A basis of such space is$$\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&0\\0&1\end{pmatrix},\begin{pmatrix}0&1\\1&0\end{pmatrix}\right\}.$$

Answer 2

We are interested in when does $T(A)=0$ which is equivalent to $A^T-A=0$, that is $$A^T=A.$$

That is it is the set of symmetric matrices.

Let $e_i$ be the standard unit vector.

A possible basis is $$\{ e_ie_j^T+e_je_i^T: 1 \le i \le j \le n\}$$

Comment about your working:

I am not sure if the problem is asking particularly about the $2$ by $2$ case. If it is, then we need $b-c=0$ and there is no constraint on $a$ and $d$.