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**tttcomrader** Let V be an inner product space, let $\displaystyle S = \{ v_{1},v_{2},...,v_{n} \} $ be an orthonormal subset of V. Prove that for any $\displaystyle x \in V $, we have $\displaystyle ||x||^2 \geq \sum ^{n} _{i=1} |<x,v_{i}>|^2 $

proof so far.

by a theorem, there exist unique vectors u in span(S) and v in the orthogonal complement of span(S) such that x = u + v.

Now, since u is orth. to v, meaning $\displaystyle ||x||^2 = ||u||^2 + ||v||^2 $

Since $\displaystyle u \in \ span{S} $, write $\displaystyle u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n} $

then $\displaystyle x = u + v = v + \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n} $

Since they are all orthogonal to one another, we can write $\displaystyle ||x||^2 = ||v||^2 + || \alpha _{1} v_{1} ||^2 + || \alpha _{2} v_{2} ||^2 + . . . + || \alpha _{n} v_{n} ||^2 $

Now, I'm a bit stuck... Am I at least doing this right so far?