Show that $W \cap W^\perp = \{0\}$ and $W + W^\perp = R^n$
I know that $W^\perp$ contains vectors perpendicular to all vectors in $W$, and that means $W$ and $W^\perp$ are linearly independent, but I have no idea where to start the proof... Can somebody give me some idea of the proof sketch? Thank you all in advance for the responses.
On
suppose that $H$ is a Hilbert space and $X$ is a closed subset of $H$. we want to show that $X+X^{\bot}=H$ and $X \cap X^{\bot}=\emptyset$. Suppose that $h \in H$ then there is $x \in X$ such that $\Vert h-x\Vert=$$inf${$\Vert h-p\Vert$ :$p \in X$}. ((Because the function $f(p)=||h-p||$ is continuous on $X$ so $f$ achieves $sup ,inf$ in $X$.)) Now Let $x^{\bot}=h-x$,we will show that $x^\bot$$\bot X$. For each$e\gt0$ we have $$||x^\bot||^2=<x^\bot,x^\bot>=<h-x,h-x>=||h-x||^2 \leq ||h-(x+ez)||^2=||x^\bot+ez||^2 \leq ||x\bot||^2+||ez||^2-2Re<x^\bot,ez>$$ so we have $$2Re<x^\bot,z> \leq e||z||^2$$ Since $e$ is arbitrary then $Re<x^\bot,z> \leq 0$ for each $z \in X$. Since $X$ is linear subspace then $<x^\bot,z>=0$,so $x^\bot\in X^\bot$.
HINT
For the first one we have
$$v\in W \cap W^\perp \implies v\cdot v=0 \implies v=0$$
For the second one let $\{v_1,v_2,\ldots,v_k\}$ a basis for $W$ then we can extend that to a basis for $R^n$ by $\{u_1,u_2,\ldots,u_r\}$ with $k+r=n$ and we need to show that the last one is a basis for $W^\perp$.