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Suppose $v_1,...,v_n \in V$ are nonzero, mutually orthogonal elements of an inner product space V. Then $v_1,...,v_n$ form an orthogonal basis for their span W = $span(v_1,...,v_n )\subset V$, which is therefore a subspace of dimension n = dimW. In particular, if dimV = n, then $v_1,...,v_n$ form a orthogonal basis for V.

How will I be able to prove this theorem?

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Well they are already orthogonal and they span, so what else do you need to show? If $c_1 v_1 + \cdots + c_n v_n = 0$, consider what happens when you take the inner product of both sides with $v_i$. –  Michael Joyce Oct 23 '12 at 5:59
    
I have to show that they are linearly independent? So, I must show that $c_i = <u,v_i>$ ? –  diimension Oct 23 '12 at 6:02
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If $u = c_1 v_1 + \cdots + c_n v_n$ and if the $v_i$ were orthonormal, then yes, you would show that $c_i = \langle u, v_i \rangle = 0$. But you don't need the $v_i$'s to be orthonormal to prove that they are linearly independent; you just need they are orthogonal to each other. –  Michael Joyce Oct 23 '12 at 6:20
    
Thank you very much! –  diimension Oct 23 '12 at 6:28
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2 Answers 2

up vote 2 down vote accepted

Since $v_1, v_2, \ldots, v_n$ is mutually orthogonal, $\{v_1,v_2,\ldots, v_n\}$ is an independent linear system in $V$. Since $dim V=n$ (equal number of elements of $\{v_1,v_2,\ldots, v_n\}$) then $\{v_1,v_2,\ldots, v_n\}$ is a basis of $V$.

To show that $\{v_1,v_2,\ldots, v_n\}$ is independent linear system we consider $$ a_1v_1+a_2v_2+\ldots+a_nv_n=0, $$ where $a_i\in \mathbb{R}$. We have $$ a_1\langle v_1, v_1\rangle + a_2\langle v_1, v_2\rangle+\ldots+a_n\langle v_1, v_n\rangle=0. $$ Hence $a_1\langle v_1, v_1\rangle=0$ due to the fact that $$ \langle v_1,v_2\rangle=\langle v_1,v_3\rangle=\ldots=\langle v_1,v_n\rangle=0. $$. Since $v_1\ne 0$, we have $a_1=0$. Argue similarly we obtain $a_2=a_3=\ldots=a_n=0$.

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Thank you very much!! So, essentially, I just have to show that they are linearly independent? –  diimension Oct 23 '12 at 6:06
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you need to show that if $c_1v_1 + ... + c_nv_n = 0$ then $c_1=...=c_n = 0$ (shows linear independence). Maybe try a dot product on the first equation

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So, I must show that $c_i = <u,v_i>$ –  diimension Oct 23 '12 at 6:05
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