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.


1 Answer 1


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, $$V \cap W = \text{span} \{(3,5,7)\}$$


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