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Office_Shredder
Office_Shredder is offline
#2
Mar3-12, 08:21 PM
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P: 4,499
If [itex] x_1,...,x_n[/itex] is a basis of V, and [itex] \left< y, x_j \right> = 0[/itex] for all j, then y=0 necessarily.

Proof: Suppose that [itex] y = a_1 x_1 +... + a_n x_n [/itex] for some numbers a1,...,an. Then if y is non-zero, [itex] \left< y, y \right> \neq 0[/itex]. But
[tex] \left< y, y \right> = \left< y, \sum a_j x_j \right> = \sum a_j \left< y, x_j \right> = 0[/tex] since every inner product in that summation is zero. So y must have been zero to begin with

I'm not where your confusion is with yi being in the orthogonal space. They haven't defined yi to be anything in particular, they're just saying, suppose they happened to pick such a vector


It may help to do an example. Let's do it in 2 dimensions: x1 = (1,0) and x2 = (1,1). Then y1 is perpendicular to everything EXCEPT for x1. In particular, y1 is perpendicular to x2 which means y1 must be of the form (a,-a) for some number a. Then <x1,y1> = a = 1 implies that a=1. So y1 = (1,-1).

<y2,x1> = 0 implies that y2 is of the form (0,c) for some number c. <y2,x2> = c, so for the inner product to be 1 it must be c=1, So y2 = (0,1)