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Complete all the details of the following argument: "Let $ S = \left\{ v_1,v_2,\ldots,v_r \right\} \subset V$ an orthonormal set with the maximum possible numbers of elements. For all $u \in V$, the vector: $$ w = u - \sum_{i=1}^r \langle u,v_i\rangle v_i $$ is orthogonal to $v_1,v_2,\ldots,v_r$. Because of $S$ is with the maximum possible numbers of elements, $w=0$, and then $S$ generates $V$, and $S$ is an orthonormal basis."

I really don't get what I should complete in the argument because for me it's almost perfect... What I've done is:

Since $w = 0$ we get that: $$ u = \sum_{i=1}^r \langle u,v_i\rangle v_i $$

and because of that, we can conclude that $u$ is always going to be a multiple of $v_i$, hence, generated by $S$.

Now we get that $S$ generates $u$ with some linear combination of it's vectors, and we get that $u$ is any vector in $V$, hence, $S$ generates $V$.

Is it just that? Am I missing any crucial part of this exercise?

Thanks!

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    Why is $w=0$? Why is $w$ orthogonal to each element in the orthonormal set? You first need to show that $w$ is indeed orthogonal to each vector in that set. If $w\neq 0$ you contradict the maximality of the given set, then you may conclude that $w$ is zero and that $w$ belong to the span of the set. You still need to show that the elements of the set are linearly indepent. Then you may conclude that they form a basis. – 2017-01-21
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    The first thing that occurred to me is proving that $w$ is orthogonal to each of $v_1,\ldots,v_r. \qquad$ – 2017-01-21
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    "that $w$ belong**s** to $\dots$", "indepen**dent**", my keyboard is seriously annoying me and I cannot edit it anymore :( – 2017-01-21
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    @Mathematician42 Thank you for answering... But I've assumed that $w=0$ because I was asked to complete details of the argument... since it said that $w=0$, i didn't need to prove it... or no? haha – 2017-01-21
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    But you didn't showed the details. The details are exactly what is said above. You cannot assume $w=0$. – 2017-01-21
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    @Mathematician42 Got it mate... So I need to prove the details, and then what I've done is going to be correct, right? – 2017-01-22

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