Prove Finite Orthogonal Set is Linearly Independent

[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

 

[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>

 

[bugatti79]

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>

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

 

[HallsofIvy]

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 $\alpha_j= 0$.

 

[blue_raver22]

.

The line comes from the fact that since u is an orthogonal set, the inner product of any two distinct vectors in the set is equal to 0. Therefore, when we take the inner product of the sum of all the vectors and any individual vector in the set, all terms except for the one with the same vector will be equal to 0. This leaves us with only the inner product of that same vector with itself, which is equal to 1 since u is an orthonormal set. Therefore, we end up with only the coefficient of that vector, which is equal to 0 since the sum of all the vectors is equal to 0. This shows that all the coefficients in the linear combination are equal to 0, proving that the set is linearly independent.