1. Proving an identity

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

No idea how to show this!

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

No idea how to show this!
Hi-Outlines of the proof.

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

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

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

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

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

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

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

Hence we are done.