Computations of matrices with respect to 2 bases. I'm really confused with the whole notion of a matrix wrt 2 bases, i've looked up numerous books and still can't grasp it so some help would be greatly appreciated here.
My Question is: suppose $(v1,v2,v3,v4)$ is a basis for $V$ , $(w1,w2,w3)$ is a basis for $W$, and that
with respect to these bases, the map $f : V → W$ corresponds to the following matrix
\begin{array}{ccc}
1 & 0 & 2 & 4 \\
3 & 1 & 0 & 6\\
2 & 2 & 1 &5 \end{array} 
$a) Find: f(v1 + 2v2).$
$b) Compute: kerf$
So what I have done so far is represented the matrix as
$F(x,y,z,w) = (x+2z+4w,3x+y+6w,2x+2y+z+5w)$
and for a) I tried to plug an arbitrary $v1$ and $v2$ into this F but it got very messy... I really don't even know how to start a).
and I found the kernel to be  $(-2,0,-1,1)$  so if you multiply this by the matrix we get the zero vector. 
So basically help with Part a would be great, thanks <3
 A: You are given a linear map $f:V\to W$ and bases 
\begin{align*}
\alpha &= \{v_1,v_2,v_3,v_4\} \\
\beta &= \{w_1,w_2,w_3\}
\end{align*}
for $V$ and $W$ respectively. You are told that the matrix of $f$ relative to these two bases is
$$
[f]_\alpha^\beta=
\begin{bmatrix}
\color{red}{1} & \color{blue}{0} & \color{orange}{2} & \color{green}{4} \\
\color{red}{3} & \color{blue}{1} & \color{orange}{0} & \color{green}{6}\\
\color{red}{2} & \color{blue}{2} & \color{orange}{1} & \color{green}{5}
\end{bmatrix}
$$
This means that
\begin{array}{rcrcrcr}
f(v_1) & = & \color{red}{1}\cdot w_1 & + & \color{red}{3}\cdot w_2 & + & \color{red}{2}\cdot w_3 \\
f(v_2) & = & \color{blue}{0}\cdot w_1 & + & \color{blue}{1}\cdot w_2 & + & \color{blue}{2}\cdot w_3 \\
f(v_3) & = & \color{orange}{2}\cdot w_1 & + & \color{orange}{0}\cdot w_2 & + & \color{orange}{1}\cdot w_3 \\
f(v_4) & = & \color{green}{4}\cdot w_1 & + & \color{green}{6}\cdot w_2 & + & \color{green}{5}\cdot w_3 
\end{array}
Part (a) asks to find $f(v_1+2v_2)$. Using the above equations and the linearity of $f$, we find that
\begin{align*}
f(v_1+2v_2)
&= f(v_1)+2f(v_2) \\
&= (w_1+3w_2+2w_3)+2\cdot(w_2+2w_3) \\
&= w_1+5w_2+6w_3
\end{align*}
Part (b) asks to find $\ker f$, which generally means find a basis for $\ker f$. To do so, we need only find a basis for $\DeclareMathOperator{Null}{Null}\Null[f]_\alpha^\beta$. Assuming you know how to do this, we find that
$$
\Null[f]_\alpha^\beta=\DeclareMathOperator{Span}{Span}\Span\left\{
\begin{bmatrix}
\color{red}{-2}\\\color{blue}{0}\\\color{orange}{-1}\\\color{green}{1}
\end{bmatrix}
\right\}
$$
This means that
\begin{align*}
\ker f
&=\Span\{\color{red}{-2}\cdot v_1+\color{blue}{0}\cdot v_2+\color{orange}{(-1)}\cdot v_3+\color{green}{1}\cdot v_4\} \\
&= \Span\{-2v_1-v_3+v_4\}
\end{align*}
