Find the kernel of T, basis for the kernel of T and the dimension of the kernel of T where $T:M_{2x2} \rightarrow M_{2x2}$

Question

Let $T:M_{2x2} \rightarrow M_{2x2}$ be a transformation defined by $T(A)=A+A^T$ and let $A= \begin{bmatrix} a & b \\ c & d \\\end{bmatrix}$

Find the kernel of T, and basis for the kernel of T and the dimension of the kernel of T.

My attempt: Set $T(A)=A+A^T=0= \begin{bmatrix} a & b \\ c & d \\\end{bmatrix}+ \begin{bmatrix} a & c \\ b & d \\\end{bmatrix} = \begin{bmatrix} 2a & b+c \\ b+c & 2d \\\end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \\\end{bmatrix}$ so, $a=d=0$ and $c=-b$ So the kernal of $T = \{\begin{bmatrix} 0 & b \\ -b & 0 \\\end{bmatrix}: b\in\Bbb R\}$

Is that correct?

So the part I'm more confused on is is finding the basis of the kernel of T. Do I just set $A= \begin{bmatrix} a & b \\ c & d \\\end{bmatrix}=0$ to get that the basis of the kernel is just $\{ \begin{bmatrix} 0 & 0 \\ 0 & 0 \\\end{bmatrix}\}$? or would the basis be $\{\begin{bmatrix} 0 \\ 0 \\\end{bmatrix}\}$

The dimension of the kernel is just how many vectors there are right?

Thanks!

Answer 1

You are correct.

Following your result, $$\text{ker}T=\text{span}\{\begin{bmatrix} 0 & 1 \\ -1 & 0 \\\end{bmatrix}\}$$

So a basis to $\text{ker}T$ would be:

$$\{\begin{bmatrix} 0 & 1 \\ -1 & 0 \\\end{bmatrix}\}$$

As we get a basis with one matrix in it (note our space is of matrices and not vectors):

$$\text{dim}\text{Ker}T=1$$