1
$\begingroup$

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

  • 0
    @DanPetersen I agree with your comment, but unfortunately when I first learned about vector spaces (and it may also be the case for the OP), *everything* was defined over $\mathbb{R}$, and so the inner product was presented as a natural construction on a vector space, instead of giving rise to a different class of algebraic objects.2012-03-23

1 Answers 1

8

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.

  • 0
    What about inverses?2017-11-12