Finding an orthonormal basis for subspace intersection, $W^{\perp}\cap V$

I am stuck trying to find an orthonormal basis for $W^{\perp}\cap V$.

I'm given V = span$\{v_1,v_2,v_3\}$ and that

$$v_1= \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix} v_2= \begin{bmatrix} 1 \\ -1 \\ 2 \\ -1 \end{bmatrix} v_3= \begin{bmatrix} 2 \\ 1 \\ 1 \\ -1 \end{bmatrix}$$

I'm also given that W = span$\{w_1,w_2\}$ and that $$w_1= \begin{bmatrix} 1 \\ 1 \\ -1 \\ -1 \end{bmatrix} w_2= \begin{bmatrix} 0 \\ 1 \\ 2 \\ -1 \end{bmatrix}$$

Now I've calculated a basis for $W^{\perp}$, which I find to be consisting of two vectors, $$y_1= \begin{bmatrix} 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} y_2= \begin{bmatrix} 3 \\ -2 \\ 1 \\ 0 \end{bmatrix}$$

I'm now uncertain how to figure determine a basis for $W^{\perp}\cap V$. Once that is complete I will be able to find an orthonormal basis (Gram Schmidt if necessary and norm the vectors to unit vectors). Where do I go from here?

• The intersection consists of vectors that are elements of both $W^\perp$ and $V$, i.e., those vectors that are linear combinations of both $\{v_1,v_2,v_3\}$ and $\{y_1,y_2\}$. Does this suggest a way to proceed? – amd May 17 '16 at 20:53
• This does make a lot of sense and feels like an 'aha!' moment. Thank you! – econom May 17 '16 at 21:38

You want to compute a basis for $W^\perp\cap V$ where $W^\perp=\DeclareMathOperator{Span}{Span}\Span\{y_1,y_2\}$ and $V=\Span\{v_1,v_2,v_3\}$. To do so, form a matrix $$A= \begin{bmatrix} y_1 & y_2 & \mid & v_1 & v_2 & v_3 \end{bmatrix}$$ In our case $$A= \left[ \begin{array}{rr|rrr} 0 & 3 & 1 & 1 & 2 \\ 1 & -2 & 1 & -1 & 1 \\ 0 & 1 & 0 & 2 & 1 \\ 1 & 0 & 1 & -1 & -1 \end{array} \right]$$ We can determine relations between $\{y_1,y_2\}$ and $\{v_1,v_2,v_3\}$ by row reducing $A$). In this case. $$\DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rr|rrr} 1 & 0 & 0 & 0 & -4 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 1 \end{array}\right]$$ The number of free columns in $\rref{A}$ is the dimension of $W^\perp\cap V$. Each free column introduces a relation. In this case $\dim W^\perp\cap V=1$. The relation is $$-4\,y_1-y_2+4\,v_1+v_2=v_3$$ which can be re-written as $$4\,v_1+v_2-v_3=4\,y_1+y_2=(3,2,1,4)$$ Thus $W^\perp\cap V=\Span\{(3,2,1,4)\}$.
• Wow I'd not considered that we're just looking to find a vector(s) that is a linear combination of both $W^{\perp}$ and $V$. This does make sense. thank you! – econom May 17 '16 at 21:37