Q. Let $U$ be the vector subspace of $\Bbb R^5$ generated by $\{(1,3,-3,-1,-4),(1,4,-1,-2,-2),(2,9,0,-5,-2)\}$, and let $V$ be the vector subspace of $\Bbb R^5$ generated by $\{(1,6,2,-2,3),(2,8,-1,-6,-5),(1,3,-1,-5,-6)\}$. What is the vector space dimension of $U \cap V$?
I found that the basis of $U$ is just $\{(1,3,-3,-1,-4),(1,4,-1,-2,-2)\}$ and that of $V$ is $\{(1,6,2,-2,3),(2,8,-1,-6,-5),(1,3,-1,-5,-6)\}$. This means $\dim U=2$ and $\dim V=3$. I know that the dimension of $U \cap V$ is either $0,1$ or $2$ since $\dim (U \cap V) \le \dim U$.
Attempt : Using the method of top answer from here, I got this equation
$a(1,3,-3,-1,-4)+b(1,4,-1,-2,-2)-x(1,6,2,-2,3)-y(2,8,-1,-6,-5)-z(1,3,-1,-5,-6)=0$
then solving the arising system of linear equations and reducing to row-echelon form I get the following system of linear equations : $$ \left\{ \begin{array}{c} a+b-x-2y+z=0 \\ b-3x-2y=0 \\ x-y-2z=0 \end{array} \right. $$
Which gives me null space as $[-2y-5z \;\;\; 5y+6z \;\;\; y+2z \;\;\; y \;\;\; z]^T$.
Setting $y=-\frac 65$ and $z=1$ we get a vector in null space as $(-\frac {13}5,0,\frac 45,-\frac 65,1)$
Hence our $\textbf {v}=-\frac {13}5 (1,3,-3,-1,-4)=\frac 45(1,6,2,-2,3)+\frac {-6}5 (2,8,-1,-6,-5)+(1,3,-1,-5,-6)$.
Does this mean that $\{(1,3,-3,-1,-4)\}$ is a basis of $U \cap V$ implying that $\dim (U \cap V)=1$? Have I done it right through out? Thanks.
PS : How I set $y=-\frac 65,z=1$ ? I let $5y+6z=0 \; \text{in null space} \; \Rightarrow y=-\frac {6z}5$. Which gives me $[-\frac {13z}5 \;\;\; 0 \;\;\; \frac {4z}5 \;\;\; -\frac {6z}5 \;\;\; z]$. Then I took $z=1$.
If I vary $z$ all over the $\Bbb R$ then,
$U \cap V=\{-\frac {13z}5 (1,3,-3,-1,-4)\ : z \in \Bbb R\}$.