Proving an identity

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.