# How to find the basis of intersection of subspaces $V$ and $W$?

The text of the problem:

Find the basis for the sum and the intersection of subspaces $$V$$ and $$W$$:

$$V=\left\{v_1=\begin{pmatrix} 2 & 1 & 0 \end{pmatrix},v_2=\begin{pmatrix} 1 & 2 & 3 \end{pmatrix},v_3=\begin{pmatrix} -5 & -2 & 1 \end{pmatrix}\right\}$$

$$W=\left\{w_1=\begin{pmatrix} 1 & 1 & 2 \end{pmatrix},w_2=\begin{pmatrix} -1 & 3 & 0 \end{pmatrix},w_3=\begin{pmatrix} 2 & 0 & 3 \end{pmatrix}\right\}$$.

What I did do:

First I'd formed the matrix $$M_V$$ and found the row echelon form of the matrix where it shows that there are two independent vectors in $$V$$, i.e. dim $$V$$=2 and formed the set $$V'=\{v_1,v_2\}$$. Similiarly, I found that dim $$W$$=2, formed $$W'=\{w_1,w_2\}$$ and then formed the basis $$B_{V+W}=\{v_1,v_2,w_1\}$$. Again, I checked if the vectors in it are linearly independant and found that indeed $$\textrm{dim } (V+W)$$=3.

After this, I get kinda stuck. I know that

dim $$V\cap W$$=dim $$V$$+dim $$W$$-dim $$(V+W)=1$$

i.e. the basis $$B_{V\cap W}$$ only has one element $$x$$ such that $$x\in V$$ and $$x\in W$$. Therefore, we have $$x=\sum_{i=1}^3\alpha_iv_i=\sum_{j=1}^3\beta_jw_j,$$ so I formed a set of equations $$2\alpha_1+\alpha_2-5\alpha_3=\beta_1-\beta_2+2\beta_3$$ $$\alpha_1+2\alpha_2-2\alpha_3=\beta_1+3\beta_2$$ $$2\alpha_2+\alpha_3=2\beta_1+3\beta_3,$$ and I have no clue where to go from here. Any help is appreciated.

Observe that $$V$$ is the plane $$x-2y+z=0$$ while $$W$$ is the plane $$x-2y+z=0$$. Finding $$V \cap W$$ means identifying the intersection of these two planes; this amounts to solving the system of equations $$x-2y+z=0$$ $$-3x-y+2z=0$$ By setting up and solving an augmented system, we see it has infinitely many solutions, and they are $$(x,y,z)=(3,5,7)t: t\in \mathbb{R}$$ In other words, $$V \cap W = \text{span} \{(3,5,7)\}$$