Vrftsjtryk

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

Consider,
$$c_1v_1+c_2v_2+ dotsb +c_nv_n =mathbf{0}.$$
Now take the inner product with vector $v_k$, to get
$$c_1 langle v_1, v_k rangle +c_2langle v_2, v_k rangle+ dotsb +c_klangle v_k, v_k rangle + dotsb +c_nlangle v_n, v_k rangle=0.$$
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.

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 $V$.

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.