Finding basis of kernel of a linear transformation I was trying to do my Linear algebra HW, and I was stuck on a question. The question is the one shown below:

I was able to do part (a), but I am having trouble in doing part (b) and (c). For part (b), I assumed an arbitrary matrix $ \
   A=
  \left[ {\begin{array}{cc}
   a & b \\
   c & d \\
  \end{array} } \right]
$, and applied the linear transformation F(A), which gave me $ \
   F(A)=
  \left[ {\begin{array}{cc}
   -a+2 b \\
   -c+2d \\
  \end{array} } \right]$. In order for it be a kernel, it must equal the zero vector, thus a=2b, c=2d. Thus $ \
   A=
  \left[ {\begin{array}{cc}
   2b & b \\
   2d & d \\
  \end{array} } \right]$
But does that mean that the ker F= {$ \
  \left[ {\begin{array}{cc}
   2 \\
   0 \\
  \end{array} } \right]$, $ \
  \left[ {\begin{array}{cc}
   0 \\
   1 \\
  \end{array} } \right]$} as these will span the whole of $R^2$ and this is what I get after separating b and d vectors. Am I doing something wrong here.
Also, how to find a coordinate vector [B] with respect to the basis that I got (or will get if I am wrong). Any help will be greatly appreciated.
 A: $\text{ker}F$ is a space of matrices here, so you need to find a basis consisting of matrices for this subspace. First letting $b=1$ and $d=0$, then the opposite, we obtain the basis
$$\left\{ \left[ {\begin{array}{cc}
   2 & 1\\
   0 & 0\\
  \end{array} } \right], \left[ {\begin{array}{cc}
   0 & 0\\
   2 & 1\\
  \end{array} } \right]\right\}$$
The coordinate vector of the matrix in the question is the vector $(a,b)$ such that $\left[ {\begin{array}{cc}
   2 & 1\\
   4 & 2\\
  \end{array} } \right]=a\left[ {\begin{array}{cc}
   2 & 1\\
   0 & 0\\
  \end{array} } \right]+b\left[ {\begin{array}{cc}
   0 & 0\\
   2 & 1\\
  \end{array} } \right]$.
A: You should have $Av=0$ and if $A$ is equal the following matrix:
$$\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
$$
Then $a=2b$ and $ c=2d$. Now $A$ is the following one:
$$\begin{bmatrix}
2b & b \\
2d & d \\
\end{bmatrix}
$$
and its basis is:
$$\left\{\begin{bmatrix}
2 & 1 \\
0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
2 & 1 \\
\end{bmatrix}\right\}$$
