Factor Modules/Vector Spaces and its basis with canonical mappings I'm having trouble with factor modules now.
Well, specifically the following question from a past paper.
Q. T=$R^2$ i.e. the real plane, and define $f:T$->$T$ with respect to the standard basis which has the 2 by 2 matrix $$A=\left( \begin{matrix} -6 & -9 \\ 4 & 6 \end{matrix} \right).$$
Give the basis $C$ for T/kerf and write the matrix for the mapping T/kerf ->T induced by $f$, with respect to the basis $C$ and the basis $B=(-3e_1+2e_2, 2e_1-e_2)$ where $e_1, e_2$ ae the standard basis for $T$.
I figured $kerf$ has to be ${v \in T; v=k(\frac{-3}{2} ; 1)}$ for some real $k$.
(Am I correct?)
So, if $u,v \in T$ and if $u+kerf = v+kerf$ then $u-v \in kerf$.
I can equate $u-v = k(\frac{-3}{2} ; 1)$ but to be honest, I don't see this leading me anywhere. My exam's tomorrow and I don't really have so much time to spend on one question right now...Can anyone please please please explain how I should think of this problem??
I might be asking much but with steps would be more than super appreciated...
My textbook doesn't even have examples like this and I am absolutely in the dark....
 A: You are correct about what the kernel is. 
When you say "Give the basis $C$..." I assume you mean "Give a basis $C$...". I.e. you have to find a basis. 
$T/ \text{ker} f$ is one-dimensional since $\dim(T/S) = \dim(T) - \dim(S)$. Any nonzero vector of a one-dimensional space constitutes a basis. (I.e. pick any vector $v$ not in $\ker f$ and $v + \ker f$ is a basis for $T/\ker f$.
The image of the map $T / \ker f \to T$ is one-dimensional also, so it consists of multiples of a single vector (in this case it consists of scalar multiples of  $(-3,2)$, say).  Write $(-3,2)$ as a linear combination of the vectors of the basis $B$. This is equivalent to solving 
$$\left( \begin{matrix} -3 & 2 \\ 2 & -1\end{matrix} \right)\left( \begin{matrix} x \\ y  \end{matrix} \right) = \left( \begin{matrix} -3 \\ 2\end{matrix} \right).$$
If you solve the above (you should be able to solve it without doing any work or writing anything down), and you pick as your basis for $T/ \ker f$ a vector $v$ such that $f(v) = (-3,2)$, then your matrix will be 
$$\left( \begin{matrix} x \\ y\end{matrix} \right).$$ 
