Given Information
We are given that $W$ is a subspace of $R^n$ with an orthogonal basis $\left\{ \mathbf { w } _ { 1 } , \ldots , \mathbf { w } _ { p } \right\}$ and let $\left\{ \mathbf { v } _ { 1 } , \ldots , \mathbf { v } _ { q } \right\}$ be an orthogonal basis for $W^\perp$. We have to explain:
(a) Why the set $\left\{ \mathbf { w } _ { 1 } , \ldots , \mathbf { w } _ { p } \right\}$ is an orthogonal set.
(b) Why the set $\left\{ \mathbf { w } _ { 1 } , \ldots , \mathbf { w } _ { p } \right\}$ span $R^n$.
(c) Show that $\operatorname { dim } W + \operatorname { dim } W ^ { \perp } = n$

Step-1: (a)
As given in the question, the vectors $\mathbf { w } _ { 1 } , \cdots , \mathbf { w } _ { p }$ are pair wise orthogonal, and the vectors $\mathbf { v } _ { 1 } , \cdots , \mathbf { v } _ { p }$ are pair wise orthogonal.
A vector x is in $W^\perp$. if x is orthogonal to every vector in a set spans W.

Since, $w_i$ is in $W$ and $v_j$ is in $W^\perp$ Thus, \[ \vec { W } _ { i } \vec { V } _ { j } = 0 \] Therefore,

The $\left\{ \mathbf { w } _ { 1 } , \cdots , \mathbf { w } _ { p } , \mathbf { v } _ { 1 } , \cdots , \mathbf { v } _ { q } \right\}$ forms an orthogonal set.

Step-2: (b)
By the orthogonal decomposition theorem, each y in $R^n$ can be written as: \[ \mathbf { y } = \hat { \mathbf { y } } + \mathbf { z } = c _ { 1 } \mathbf { w } _ { 1 } + \cdots + c _ { p } \mathbf { w } _ { p } + d _ { 1 } \mathbf { v } _ { 1 } + \cdots + d _ { q } \mathbf { v } _ { q } \] Therefore,

The $\left\{ \mathbf { w } _ { 1 } , \cdots , \mathbf { w } _ { p } , \mathbf { v } _ { 1 } , \cdots , \mathbf { v } _ { q } \right\}$ spans $R^n$

Step-3: (c)
From part (a) the set is linearly independent and from (b) it spans $R^n$. Hence the set forms a basis for $R^n$.
Therefore,

The $\left\{ \mathbf { w } _ { 1 } , \cdots , \mathbf { w } _ { p } , \mathbf { v } _ { 1 } , \cdots , \mathbf { v } _ { q } \right\}$ is a basis for $R^n$