Composition of linear transformations different

In $\mathbb{R} ^3$ a base $A=\{\alpha_1,\alpha_2,\alpha_3\}$ and in $\mathbb{R}^2$, $B=\{\beta_1,\beta_2\}$ are given, where $\alpha_1=[1,1,1],\alpha_2=[1,1,0],\alpha_3=[1,0,0]$ and $\beta_1=[1,1],\beta_2=[3,2]$.

Let a linear transformation $\varphi : \mathbb{R}^3\rightarrow \mathbb{R}^2$ be described by a matrix $$M(\varphi)_A^B=\begin{pmatrix} 12 & 8 & 4 \\ -3 & -2 & -1 \end{pmatrix},$$ and let $\psi : \mathbb{R}^4\rightarrow \mathbb{R}^3$ be given by an equation $$\psi([x_1,x_2,x_3,x_4])=[x_1+2x_2+3x_3+x_4,2x_1+x_2+2x_3+x_4,-x_1+x_2+x_3].$$

a) Find the formula for $\varphi \circ \psi : \mathbb{R}^4 \rightarrow \mathbb{R}^2$;

b) Find base of $V\cap W$, where $W=\ker \psi$ and $V={\rm im}\psi$ are subspaces of $\mathbb{R}^3$.

I have utterly no idea how to start a), b). I think it is easier but without a) i do not know for certain. How to approach this question?

You can find the matrix associated to $\varphi$ with respect to the standard bases; this is $$M(\varphi)=\begin{bmatrix} 1 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 12 & 8 & 4 \\ -3 & -2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}^{-1}= \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \end{bmatrix}$$ (do the computations).
The matrix associated to $\psi$ with respect to the standard bases is $$M(\psi)=\begin{bmatrix} 1 & 2 & 3 & 1 \\ 2 & 1 & 2 & 1 \\ -1 & 1 & 1 & 0 \end{bmatrix}$$ so the matrix of $\varphi\circ\psi$ is the product.
From the matrix it's easy to represent $\varphi\circ\psi$.
For part (b), I believe you want to find $$\operatorname{im}\psi\cap\ker\varphi$$ because $\ker\psi$ is not a subspace of $\mathbb{R}^3$.
A basis of $\operatorname{im}\psi$ is given by the first two columns of $M(\psi)$ (why?); when does a linear combination of these two vectors belong to $\ker\varphi$?
• You're right, my mistake it should be $\ker \varphi$ – kurkowski Jan 8 '15 at 16:02