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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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    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.2012-10-23

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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?2012-10-23
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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 = $2012-10-23