2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – amd May 17 '16 at 20:53
  • $\begingroup$ This does make a lot of sense and feels like an 'aha!' moment. Thank you! $\endgroup$ – econom May 17 '16 at 21:38
1
$\begingroup$

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 $W^\perp\cap V=\Span\{(3,2,1,4)\}$.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – econom May 17 '16 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.