Let be $V=\{ \vec{x} \in \mathbb{R}^4 | x+y+z+t=0 \}$ and $W=span(1,2,3,4)^T$ two linear subspaces of $\mathbb{R}^4$.
$E=\{ f \in End(\mathbb{R}^4) | f(V) \subseteq W\}$ is a linear subspace of $End(\mathbb{R}^4)$ ? What is its dimension?
And $E'=\{ f \in End(\mathbb{R}^4) | f(V) = W\}$ is a linear subspace of $End(\mathbb{R}^4)$ ?
MY IDEA
$V=span\{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}$ so to find the representative matrix of $f$ I need to find a matrix $M$ 4x4 such that:
$M[(1,0,0,-1)] \in (1,2,3,4)$
$M[(0,1,0,-1)] \in (1,2,3,4)$
$M[(0,0,1,-1)] \in (1,2,3,4)$ But how to find it? (Thanks to $M$ I can find the dimension of $E$)
The dimension of $V$ is $3$ and the dimension of $W$ is $1$ and this is all you need to know.
Extend a basis $\{v_1,v_2,v_3\}$ of $V$ to a basis $\mathscr{B}=\{v_1,v_2,v_3,v_4\}$ of $\mathbb{R}^4$ and a basis $\{w_1\}$ of $W$ to a basis $\mathscr{C}=\{w_1,w_2,w_3,w_4\}$ of $\mathbb{R}^4$.
A linear map $f\colon\mathbb{R}^4\to\mathbb{R}^4$ will belong to $E$ if and only if its matrix with respect to $\mathscr{B}$ and $\mathscr{C}$ is of the form $$ \begin{bmatrix} a & b & c & p \\ 0 & 0 & 0 & q \\ 0 & 0 & 0 & r \\ 0 & 0 & 0 & s \end{bmatrix} $$ This is because $f(v_1)$, $f(v_2)$ and $f(v_3)$ are required to be scalar multiples of $w_1$, while $f(v_4)$ can be any linear combination of $\mathscr{C}$.
Can you now tell the dimension of $E$?
For the other question, does the zero endomorphism belong to $E'$?