Exist $$U:=Span(\overbrace{e_1}^{u_1},\overbrace{e_2+e_3}^{u_2})\leq \mathbb{R}^4$$$$W:=Span(\overbrace{e_1+e_4}^{w_1},\overbrace{e_2+e_4}^{w_2})\leq \mathbb{R}^4$$
$T: V \to V$ is a projection on $U$ parallel on $W$.
Calculate $[T]_E$ when $E$ is the standard basis.
My solution:
$dim(U)=2, dim(W)=2$.
I'll show $U\cap W=\{ 0\}.$
$a(e_1)+b(e_2+e_3)=c(e_1+e_4)+d(e_2+e_4)$ , $a,b,c,d \in \mathbb{R}.$ $(a,b,b,0)=(c,d,0,c+d) \implies a=c , b=d=0, c=-d \implies {a=b=c=d=0}.$
Then $U\cap W=\{0\},dim(U)+dim(W)=2+2=4=dim(\mathbb{R}^4) \implies{\mathbb{R^4}=U\oplus W} $
$e_1=1\cdot u_1 \implies T(e_1)=1\cdot u_1$
$e_2=1\cdot(u_1)+0\cdot(u_2)-1\cdot(w_1)+1\cdot(w_2) \implies {T(e_2)=1\cdot(u_1)+0\cdot(u_2)}$
$e_3=-1\cdot(u_1)+1\cdot(u_2)+1\cdot(w_1)-1\cdot(w_2) \implies {T(e_3)=-1\cdot(u_1)+1\cdot(u_2)}$
$e_4=-1\cdot(u_1)+0\cdot(u_2)+1\cdot(w_1)+0\cdot(w_2) \implies {T(e_4)=-1\cdot(u_1)+0\cdot(u_2)}$
Then $[T]_E = \begin{pmatrix} 1 & 1 & -1 & -1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$
Is my solution correct ? I'd be grateful for your feedback!
You were asked to find the matrix in the standard basis. So you should express $T(e_j)$ as a linear combination of $e_1,e_2,e_3,e_4$ and not as a linear combination of the $u_i$'s to find the elements of the matrix. The matrix you found is one which maps the coordinates of a vector $v$ in the standard basis to the coordinates of $T(v)$ in the basis $u_1,u_2,w_1,w_2$ so it's not $[T]_E$. You should have
$$ T(e_1) = u_1 = e_1$$ $$ T(e_2) = u_1 = e_1 $$ $$ T(e_3) = u_2 - u_1 = -e_1 + e_2 + e_3$$ $$ T(e_4) = -u_1 = -e_1$$
So the matrix in the basis $E$ is
$$ [T]_E = \begin{bmatrix}1 & 1 & -1 & -1\\ 0 & 0 & 1 & 0 \\ 0& 0 &1 & 0\\ 0&0&0&0 \end{bmatrix}.$$