Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$W$ is a subspace of the vector space $V$. Show that $W^{\perp}$ is also a subspace of $V$.

share|improve this question
4  
What have you tried? –  Alex Becker Mar 23 '12 at 17:00
    
Use the linearity of the dot product and the fact that the zero vector is in $W^\perp$. –  David Mitra Mar 23 '12 at 17:05
    
I tried saying that a vector in $W$ is orthogonal to all vectors in $W^\perp$. But I'm not sure how to link this to $V$. –  quantum Mar 23 '12 at 17:06
3  
Just verify one by one the conditions for subspace. (i) Is the $0$-vector in the orthogonal complement? (ii) Is the sum of two vectors in the orthogonal complement also in? (iii) What about a constant times a vector in the orthogonal complement? Each verification is I think mechanical. –  André Nicolas Mar 23 '12 at 17:13
    
Yup I figured it out. Thanks for the hints guys. –  quantum Mar 23 '12 at 17:14
show 2 more comments

1 Answer

up vote 3 down vote accepted

You need to show three things:

  1. $W^\perp$ is non-empty.
  2. $W^\perp$ is closed under scalar multiplication; that is, if ${\bf v}\in W^\perp$, then $\alpha{\bf v}\in W^\perp$ for all scalars $\alpha$.

  3. $W^\perp$ is closed under vector addition; that is, if ${\bf v_1}\in W^\perp$ and ${\bf v_2}\in W^\perp$, then ${\bf v_1}+{\bf v}_2\in W^\perp$.


Recall that $\bf v$ is in $W^\perp$ if and only if ${\bf v}\cdot {\bf w}=0$ for all ${\bf w}\in W$.


Towards showing 1) holds, note (and verify) that the zero vector is in $W^\perp$.

Towards showing 3) holds, suppose that ${\bf v}_1$ and ${\bf v}_2$ are both in $W^\perp$. We have to show that the vector ${\bf v}_1+{\bf v}_2$ is in $W^\perp$; so we need to verify that $({\bf v}_1+{\bf v}_2)\cdot {\bf w}=0$ for all ${\bf w}\in W$. Towards this end, use the fact that $({\bf v}_1+{\bf v}_2)\cdot {\bf w}= {\bf v}_1\cdot {\bf w}+{\bf v}_2\cdot {\bf w} $.

I'll leave the verification that 2) holds, and the rest of the verification that 3) holds for you.

share|improve this answer
    
@BrandonCarter yes, thanks. Fixed now. –  David Mitra Mar 23 '12 at 17:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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