Representation theory and direct sum I came across the following theorem in one of the online notes regarding representation theory which I thought should have a simple proof. I am trying to prove it using basic linear algebra tools:
Theorem:
If W is a subspace of a Hermitian vector space (V, ( , )) let $W^{\perp}= \{v|<v,w>=0, \forall w\in W\}$ then $V=W\oplus W^{\perp}$ (direct sum)
(Here, V and W are finite dimensional)
I do have a rough idea as to how to go about this proof but I am not sure how to formulate it rigorously:
First just to be sure, I have to show that intersection of $W$ and  $W^{\perp} $ is trivial that is they intersect at a single point and then show $dim W^{\perp}= dim V-dim W$. Maybe we also need gram matrix to go about this but I am not sure. Any help is appreciated
 A: To show that $V = W \oplus W^\perp$, we need to show that $W\cap W^\perp = \{\vec{0}\}$ and $V = W + W^\perp$.
If we want a some vector $\vec{v}$ in both $\vec{v}\in W$ and $\vec{v}\in W^\perp$, then it follows that 
$$
<\vec{v},\vec{w}> = 0, \forall w\in W~~~~~\text{and in particular, since $\vec{v}\in W$}~~~~~<\vec{v},\vec{v}> = 0
$$
It is a basic property of any inner product that this only occurs if $\vec{v} = \vec{0}$. Thus  $W\cap W^\perp = \{\vec{0}\}$.
Next, suppose $\beta = \{\vec{u}_1,\vec{u}_2,..\vec{u}_m\}$ is an orthonormal basis for $W$. This basis can be extended to $\gamma = \beta\cup\{\vec{u}_{m+1},..\vec{u}_n\}$ an orthonormal basis for $V$. We claim that $\delta = \{\vec{u}_{m+1},..\vec{u}_n\}$ is a basis for $W^\perp$. 
It is evident that the vectors in $\delta$ are linearly independent. Now, we show they span $W^\perp$. Consider any vector $\vec{v}\in W^\perp$. Since $\vec{v}\in V$ also, we can say that
$$
\vec{v} = a_1\vec{u}_1 + a_2\vec{u}_2 + \cdots + a_n\vec{u}_n
$$
Since $\vec{v}\in W^\perp$ it follows that
$$
<\vec{v},\vec{u}_i> = <a_1\vec{u}_1 + a_2\vec{u}_2 + \cdots + a_n\vec{u}_n,\vec{u}_i> = a_i<\vec{u}_i,\vec{u}_i> = a_i = 0~~~~1 \leq i \leq m
$$
Thus $\vec{v} = a_{m+1}\vec{u}_{m+1} + \cdots + a_n\vec{u}_n$ so $W^\perp \subseteq span(\delta)$. Observe that $\vec{u}_i\in W^\perp$ if $m<i\leq n$, since $\beta$ is orthonormal
$$
\begin{eqnarray}
<\vec{u}_i,\vec{w}> &=& <\vec{u}_i,a_1\vec{u}_1 + a_2\vec{u}_2 + \cdots + a_m\vec{u}_m>\\ &=& a_1<\vec{u}_i,\vec{u}_1> + \cdots + a_m<\vec{u}_i,\vec{u}_m> \\&=& 0 +\cdots + 0 \\&=& 0\end{eqnarray}
$$
Thus the inner product of any linear combination of $\vec{u}_m...\vec{u}_n$ with some vector $\vec{w}\in W$ will also be zero, so $span(\delta) \subseteq W^\perp$. It follows that $span(\delta) = W^\perp$, and $\delta$ is a basis for $W^\perp$.
It is evident that $V = W + W^\perp$, since the union of the bases for $W$ and $W^\perp$ is a basis for $V$: $\beta = \gamma \cup \delta$.
We have satisfied $W\cap W^\perp = \{\vec{0}\}$ and $V = W + W^\perp$, so $V = W \oplus W^\perp$.
NOTE: I realize this proof is far from neat and inelegantly unconcise. I am open to suggestions.
