Mrthfx

I need help with determining the basis of $U_1 cap U_2$ in the following problem:

Let $V=mathbb{R}^4$. ${U_1} = left{ {left( {begin{array}{*{20}{c}}
{{x_1}} \
{{x_2}} \
{{x_3}} \
{{x_4}}
end{array}} right)left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} right.} right}$ and $U_2=leftlangle {left( {begin{array}{*{20}{c}}
1 \
1 \
0 \
3
end{array}} right),left( {begin{array}{*{20}{c}}
0 \
{ - 1} \
0 \
1
end{array}} right)} rightrangle$.

If $U_1$ is a subspace of $V$, determine a basis of $U_1 cap U_2$.

My attempt:

I know that ${U_2} = left{ {left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)left| {lambda ,mu in mathbb{R}} right.} right}$, and that the next step is that I should choose an element in $U_1$ and in $U_2$, e.g. Let $w in {U_1}$ and let $w in {U_2}$. Then we know that $w$ is of the form $w = left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)$, but I'm not sure what the procedure is from there.

The next step is to note thatbegin{align}U_1cap U_2&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,lambda-(lambda-mu)-3(3lambda+mu)=0right}\&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,9lambda+2mu=0right}\&=left{begin{pmatrix}lambda\frac{11}2lambda\0\-frac32lambdaend{pmatrix},middle|,lambdainmathbb{R}right}.end{align}Can you take it from here?

$U_1$ is the set of vector such that $begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} x_1 \x_2 \x_3 \x_4end{bmatrix} = 0$

$begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 1 \1 \0 \3end{bmatrix} = -9\
begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = -2$

$2begin{bmatrix} 1 \1 \0 \3end{bmatrix} - 9begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = begin{bmatrix} 2 \11 \0 \-3end{bmatrix}$

If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 qquad & qquad b = x_4-3x_1 Rightarrow x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 & x_3=0. $$
If $(x_1,x_2,x_3,x_4) in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) in U_1 cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ left[ begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \ 4 & -1 & 0 & -1 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim
left[ begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim left[ begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 cap U_2 = text{span}(-2,-11,0,3). $$