kernel and image of a map help The question states:

a) Suppose that $U = V = M_{2,2} (\mathbb R)$ is the vector space of $2 × 2$ matrices over $\mathbb R $ Let $\phi$ be defined by $\phi(A) = A - A^T $ for each $A \in M_{2,2} \mathbb R$. Determine
$\ker \phi $ and state its dimension.


b) Determine $\operatorname{im}\phi$ and its dimension.

So my thoughts are we use $\ker \phi  = \left \{ f \in U \mid \phi(f) = 0 \right \}$ and then we follow all the steps to show that  $$\ker \phi = \left \langle \begin{pmatrix}
1  & 0\\ 
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\ 
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0\\ 
0 & 1
\end{pmatrix}\right \rangle
$$
and we can conclude that $\dim \ker \phi = 3 $ and so by Rank and Nullity Thm, we say $\dim\operatorname{im}\phi = 1$.
My question is that is this correct, this process to get that answer, and I'm just wondering what $\operatorname{im}\phi$ would be?
Thank you in advance.
Edit: How I got to the kernel:
$$f = \begin{pmatrix}
a & b\\ 
c & d
\end{pmatrix} $$
then $$f^T = \begin{pmatrix}
a & c\\ 
b & d
\end{pmatrix}$$
then $$f-f^T = \begin{pmatrix}
0 & c-b\\ 
c-b & 0
\end{pmatrix}$$
So putting into a system of eqs.
$$\left \{ b-c = 0 \right. \\
\left \{ c-b = 0 \right.$$
Which implies that b=c
So $$f = \begin{pmatrix}
a & b\\ 
c & d
\end{pmatrix} = a\begin{pmatrix}
1 & 0\\ 
0 & 0
\end{pmatrix} +b \begin{pmatrix}
0 & 1\\ 
 0& 0
\end{pmatrix} + d \begin{pmatrix}
0 & 0\\ 
 0& 1
\end{pmatrix} $$
Which allows me to come to the conclusion that :
$$ker \phi = \left \langle \begin{pmatrix}
1  & 0\\ 
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\ 
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0\\ 
0 & 1
\end{pmatrix}\right \rangle
$$

Thank you for telling me to edit


I just realised that for kernel 2nd part, it should be $$\begin{pmatrix}
0 &1 \\ 
1 & 0
\end{pmatrix}$$


The first part is correct.  It is easy to see that the kernel is the set of symmetric matrices, which for dimension $n$ has dimension $n(n+1)/2$.
Since the image is one dimensional, it suffices to find one nonzero element of it, and that will form a basis.  For instance, we have that $\begin{pmatrix}1&1\\0&1\end{pmatrix}\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, and the latter matrix spans the image.  Notice it is antisymmetric.
