Let $V$ be an inner product space. suppose $S=\{v_1,v_2, \ldots,v_n\}$ is an orthogonal set of nonzero vectors Let $V$ be an inner product space. suppose $S=\{v_1,v_2, ... v_n\}$ is an orthogonal set of nonzero vectors in $V$ such that $V = \text{Span}(S)$. Prove that $S$ is a basis for $V$.
Problem
 A: Consider,
$$c_1v_1+c_2v_2+ \dotsb +c_nv_n =\mathbf{0}.$$
Now take the inner product with vector $v_k$, to get
$$c_1 \langle v_1, v_k \rangle +c_2\langle v_2, v_k \rangle+ \dotsb +c_k\langle v_k, v_k \rangle + \dotsb +c_n\langle v_n, v_k \rangle=0.$$
From orthogonality it follows that all terms on the left side are zero, except one term. So we get
$$c_k \langle v_k, v_k \rangle =c_k\|v_k\|^2=0.$$
But $v_k$ is a nonzero vector. Thus $c_k=0$. Likewise we can have all the coefficients as $0$. Thus the set is linearly independent and since it already spans, therefore a basis.
A: I am going to assume that $V$ is an inner product space over the real field $\Bbb R$; where the inner product is a bilinear mapping
$\langle \cdot, \cdot \rangle: V \times V \to \Bbb R \tag 1$
such that, for all $v \in V$,
$\Vert v \Vert^2 = \langle v, v \rangle \ge 0, \tag 2$
where equality holds if and only if
$v = 0; \tag 3$
then if a linear dependence existed between the elements of $S$, we would have
$\alpha_i \in \Bbb R, \; 1 \le i \le n, \tag 4$
such that
$\exists i, \; 1 \le i \le n, \alpha_i \ne 0, \tag 5$
that is, not all the $\alpha_i$ vanish, such that
$\displaystyle \sum_1^n \alpha_i v_i = 0.  \tag 6$
If we take the inner product of each side of this equation with any $v_j$, we find that
$\alpha_j \langle v_j, v_j \rangle = \displaystyle \sum_1^n \alpha_i \langle v_j, v_i \rangle = \left \langle v_j,  \displaystyle \sum_1^n \alpha_i v_i \right \rangle = \langle v_j, 0 \rangle = 0, \tag 7$
since $i \ne j$ implies
$\langle v_j, v_i \rangle = 0 \tag 8$
by the orthogonality of the members of $S$.  Since
$v_j \ne 0, \; 1 \le j \le n, \tag 9$
we have
$\langle v_j, v_j \rangle \ne 0, \tag{10}$
whence (7) yields
$\alpha_j = 0, \; 1 \le j \le n; \tag{11}$
it follows that the set $S$ is linearly independent over $\Bbb R$, and hence since
$V = \text{Span}(S), \tag{12}$
that $S$ is a basis for $V$.
A: Since the vectors $v_1,\ldots,v_n$ are orthogonal to each other, they are all linear independent. So we have $n$ linear independent vectors of the right dimension, thus $S$ is a basis for $V$.
Edit:
Why are the vectors linear independent?
First, checkout the definition here.
For any linear combination $v = \sum_{j=1}^n a_j v_j$ of the non-zero vectors $v_j$ we know that if $v = 0$ holds, then all $a_j$ must be zero to fulfill the equality (since all $v_j \neq 0$), thus fulfilling the definition of linear independence.
