# Orthogonal Sets

1. Mar 22, 2012

### bugatti79

Folks,

I am looking at my notes. Wondering where the highlighted comes from.
Prove that a finite orthogonal set is lineaarly independent

let u=(x_1,x_2,x_n) bee an orthogonal set set of vectors in an ips.
To show u is linearly independent suppose

Ʃ $\alpha_i x_i=0$ for i=1 to n

Fix any j=1 and consider <Ʃ$\alpha_i x_i, x_j$> i=1 to n

then

0=<Ʃ$\alpha_i x_i, x_j$> i=1 to n

=Ʃ<$\alpha_i x_i, x_j$> i=1 to n

=Ʃ$\alpha_i <x_i, x_j>$ i=1 to n

=$\alpha_j <x_j, x_j>$ since u is an orthonormal set

Where does this line come from? Thanks

2. Mar 22, 2012

### TwilightTulip

You are assuming the x_i's are orthogonal (i.e. <x_i,x_j>=0 if i =/= j), so the only term in the sum which is not necessarily 0 is a_j<x_j,x_j>

3. Mar 28, 2012

### bugatti79

where <x_j,x_j>=1 when i=j? Thanks in advance.

4. Mar 28, 2012

### HallsofIvy

Staff Emeritus
Not necessarily. If you were to use "orthonormal" instead of just "orthogonal", then that would be true.

However, that is not necessary to your proof. $<x_j, x_j>$ is some non-zero number. $a_j$. Divide both sides of $\alpha_j a_j= 0$ by that number to get $\alpha_j= 0$.