Proof: Sum of dimension of orthogonal complement and vector subspace 
Let $V$ be a finite dimensional real vector space with inner product $\langle \, , \rangle$ and let $W$ be a subspace of $V$. The orthogonal complement of $W$ is defined as
  $$
W^\perp= \left\{ v \in V \,:\, \langle v,w \rangle = 0 \text{ for all } w \in W \right\}.
$$
  Prove the following: $\dim W + \dim W^\perp= \dim V$.

I'm not sure how to find the relationship between number of basis vectors in $W$ and $W^\perp$.
 A: Hint:
Take a basis $w_1,\dots,w_r$ of $W$, and consider the linear forms on $V$ defined by $w_i^*:v\mapsto\langle w_i,v\rangle$.
These  linear forms are linearly independent, hence the solutions of the system of equations $w_i^*(v)=0,\ i=1,\dots r$ has codimension $r$ by the rank-nullity theorem. These solutions are precisely the orthogonal complement $\;U^{\bot}$.
A: It is sufficient to show that $V=W\oplus W^{\perp}$. If $v\in W\cap W^{\perp}$, then $\left\langle v,v\right\rangle=0$. Hence it remains to show that any vector $v\in V$ can be written as $v=w+w'$ with $w\in W$ and $w'\in W^{\perp}$.
Choose an orthonormal basis $\left\{w_1,\dots , w_k\right\}$ of $W$ and extend to an orthonormal basis $\left\{w_1,\dots,w_k,v_{k+1},\dots ,v_n\right\}$ of $V$. By definition $v_i\in W^{\perp}$ for all $n\geq i\geq k+1$. Hence any $v\in V$ can be decomposed as we needed to show.
A: Let $\beta=\{w_1,w_2,\ldots,w_k\}$ and $\gamma=\{x_1,x_2,\ldots,x_m\}$ be the bases for $W$ and $W^\perp$, respectively. It suffices to show that
$$\beta\cup\gamma=\{w_1,w_2,\ldots,w_k,x_1,x_2,\ldots,x_m\}$$
is a basis for $V$.
Given $v\in V$, then it is well-known that $v=v_1+v_2$ for some $v_1\in W$ and $v_2\in W^\perp$. Also because $\beta$ and $\gamma$ are bases for $W$ and $W^\perp$, respectively, there exist scalars
$a_1,a_2,\ldots,a_k,b_1,b_2,\ldots,b_m$ such that
$v_1=\displaystyle\sum_{i=1}^ka_iw_i$ and $v_2=\displaystyle\sum_{j=1}^mb_jx_j$. Therefore
$$v=v_1+v_2=\sum_{i=1}^ka_iw_i+\sum_{j=1}^mb_jx_j,$$
which follows that $\beta\cup\gamma$ generates $V$. Next, we show that 
$\beta\cup\gamma$ is linearly independent. Given 
$c_1,c_2,\ldots,c_k,d_1,d_2,\ldots,d_m$ such that
$\displaystyle\sum_{i=1}^kc_iw_i+\sum_{j=1}^md_jx_j={\it 0}$, then
$\displaystyle\sum_{i=1}^kc_iw_i=-\sum_{j=1}^md_jx_j$. It follows that 
$$\sum_{i=1}^kc_iw_i\in W\cap W^\perp\quad\mbox{and}\quad
\sum_{j=1}^md_jx_j\in W\cap W^\perp.$$
But since $W\cap W^\perp=\{{\it 0}\,\}$ (gievn $x\in W\cap W^\perp$,
we have $\langle x,x\rangle=0$ and thus $x={\it 0}\,$), we have
$\displaystyle\sum_{i=1}^kc_iw_i=\sum_{j=1}^md_jx_j={\it 0}$. Therefore
$c_i=0$ and $d_j=0$ for each $i,j$ becasue $\beta$ and $\gamma$ are bases
for $W$ and $W^\perp$, respectively. Hence we conclude that $\beta\cup\gamma$ is linearly independent.
A: There is something missing in @Solumilkyu 's answer.
In order to show that $W$ and $W^\perp$ have bases, we must show that $W, W^\perp \neq \{\vec{0}\}$, which cannot always be satisfied.
In fact, if $W=V$, then from $W \cap W^\perp=\{\vec{0}\}$ (To prove this, let $\vec{x} \in W \cap W^\perp$. Then $<\vec{x},\vec{x}>=0 \to \vec{x}=\vec{0}$) we can conclude that $W^\perp=\{\vec{0}\}$. Then $\dim W+\dim W^\perp=\dim W$. If $W\neq V$, then from the fact that $v \in V$ implies that $v=v_1+v_2$ for $v_1 \in W \wedge v_2 \in W^\perp$, we can conclude that $W^\perp \neq \{\vec{0}\}$ just by setting $v \in V \setminus W \wedge v \neq \{\vec{0}\}$ and $v_1 = \vec{0}$.
On the other hand, if $W = \{\vec{0}\}$, then obviously $W^\perp = V$.
A: I have some qualms with @Solumilkyu’s answer. To prove that that a set of vectors is indeed a basis, one needs to prove prove both, spanning property and the independence.  @Solumilkyu has demonstrated $\beta \cup \gamma$ is linearly independent, but has very conveniently assumed the spanning property.
To answer the question, though its quite old, one can use the rank-nullity theorem, coupled with the fact that $\text{rank}(A) = \text{rank}(A^{T})$ and $N(A^{T}) = V^{\bot}$.
