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

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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
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

1 Answer 1

up vote 4 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.

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@BrandonCarter yes, thanks. Fixed now. –  David Mitra Mar 23 '12 at 17:30

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