linear algebra - Base of subspaces and intersections

Let $V$ be a vectorial space with base $[v_1,v_2,v_3,v_4]$. Let $ U \subset V $ be the subspace generated by $ u_1 = v_1-v_2+v_4 $, $ u_2 = v_3+v_4 $,$ u_3 = v_1-v_2-v_3 $ and $ W \subset V$ the subspace generted by $w_1 = v_1 - 2v_3$,$w_2 = v_2 +v_4$,$w_3 = v_1 + 2v_3-v_4$,$w_4 = 2v_1 + v_2$.

a) Find a base of the subspaces $ U,W$.

b) Find a base of $ U \cap W$, $ U + W$.

c) Find a base of a subspace $L$ such that $ U\oplus L = V$.

d) Let $ f:V\rightarrow V$ the linear application defined by $f(v_1)=v_1$,$f(v_2)=v_2$,$f(v_3)=0$,$f(v_4)=0.$ Find a base of $f(U)$,$f(W)$.

a)
I have:
$ u_1 = v_1-v_2+v_4 $, $ u_2 = v_3+v_4 $,$ u_3 = v_1-v_2-v_3 $
so
$ U = <(1,-1,0,1),(0,0,1,1),(1,-1,-1,0)>$.
The three vectors are not linearly indipendent, the second is combination of the other two so
$$ U = <(1,-1,0,1),(0,0,1,1)>.$$
For $W$ I have
$ W = <(1,0,-2,0),(0,1,0,1),(1,0,2,-1),(2,1,0,0)>$

They are not indipendent, one is combination of the others thus
$$ W = <(1,0,-2,0),(0,1,0,1),(1,0,2,-1)>$$

b) Now a base of $ U \cap W$.
$ v \in U$ then

$ v = x(1,-1,0,1)+y(0,0,1,1) = (x,-x,y,x+y)$
and $ v \in W$ then
$ v = a(1,0,-2,0)+b(0,1,0,1)+c(1,0,2,-1) = (a+c,b,-2a+2c,b-c)$
the system yields the solution

$ x = a, b = -a, y = -2a, c = 0$
Thus
$ v = (x,-x,y,x+y)=(a+c,b,-2a+2c,b-c) = (a,-a,-2a,-a)$

So a base of $ U \cap W$ is:

$ U \cap W = <(1,-1,-2,-1)>$

Now I have:
$ \dim U + \dim W = \dim U\cap W + \dim U+W $

so

$ \dim U+W = 4$
so a base of $ U+W$ is any base of $ R^4$ like
$ U+W = <(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)>$

c)
The vectors that generate $L$ must be indipendent with the vectors of $U$ but together they must be a base of $V$.
So $u_1 = v_1-v_2+v_4 $, $ u_2 = v_3+v_4 $
I can pick $ u_3 = v_4 $, $ u_4 = v_1$

in this way the vectors are linearly indipendent and they span $V$ because $ v_1 = u_4$, $v_2 = -u_1 + u_4 + u_3$, $ v_3 = u_2 -u_3$ and $ v_4 = u_3$

So $ L = <(0,0,0,1),(1,0,0,0)>$

d)$ U = $
So
$ f(U) = < f(u_1),f(u_2)> = = <(1,-1,0,0)>$

$ W = $
So

$ f(W) = < f(w_1),f(w_2),f(w_3)> = = <(1,0,0,0),(0,1,0,0)>$

I would like to know if everything is right.