# Proving an identity

• Oct 12th 2009, 01:29 PM
Krizalid
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! :D
• Oct 12th 2009, 09:14 PM
aman_cc
Quote:

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! :D

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.