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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

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closed as off-topic by Adrian Keister, Chris Custer, Brahadeesh, KReiser, Cesareo Dec 13 '18 at 8:53

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  • $\begingroup$ I edited your question to clean up the $\LaTeX$ a little bit. Cheers! $\endgroup$ – Robert Lewis Dec 2 '18 at 19:29
  • $\begingroup$ It really boils down to showing that orthogonality implies linear independence; see Anurag A's answer. $\endgroup$ – Dave Dec 2 '18 at 20:09
  • $\begingroup$ Thanks Dave got it right :) $\endgroup$ – Anas Dec 3 '18 at 4:24
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    $\begingroup$ @RobertLewis Haha thanks man! $\endgroup$ – Anas Dec 3 '18 at 4:25
  • $\begingroup$ My pleasure my friend! 😁😁😁 $\endgroup$ – Robert Lewis Dec 3 '18 at 4:31
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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.

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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 $V$.

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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.

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