# Proving an identity

• Oct 12th 2009, 01:29 PM
Krizalid
Proving an identity
Let $S,W$ be subspaces of $V$ where $V$ is a vector space with inner product and $\dim V=n,$ show that $(S\cap W)^\perp=S^\perp+W^\perp.$

No idea how to show this! :D
• Oct 12th 2009, 09:14 PM
aman_cc
Quote:

Originally Posted by Krizalid
Let $S,W$ be subspaces of $V$ where $V$ is a vector space with inner product and $\dim V=n,$ show that $(S\cap W)^\perp=S^\perp+W^\perp.$

No idea how to show this! :D

Hi-Outlines of the proof.

1. $S^\perp+W^\perp \subseteq (S\cap W)^\perp$
This is easy to show.

2. So if we show that dimension of LHS = dimension of RHS we are done

3. $S^\perp \cap W^\perp = (S + W)^\perp$
This can again be shown easily

4. dim( $W$) + dim( $W^\perp$) = dim( $V$) = $n$
We will use this result directly. This is applicable for any sub-space W in V.

5. dim( $S+W$) = dim( $S$) + dim( $W$) - dim( $S \cap W$)
We will use this result directly. This is applicable for any sub-paces S,W in V.

Let dim( $S$) = s, dim( $W$)=w, dim( $S \cap W$) = i, dim( $S+W$) = u
so, $s+w = u+i$

dim( $S^\perp+W^\perp$) = dim( $S^\perp$)+dim( $W^\perp$) - dim( $S^\perp \cap W^\perp$)
= $(n-s)+(n-w)-(n-u)$
= $(n-i)$
= dim( $(S\cap W)^\perp$)

Hence we are done.