# Homework Help: Orthogonal Basis and Inner Products

1. Mar 3, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

3. The attempt at a solutionSince A is a vector in V and since the $A_i$ form a basis, we can write A as a linear combination of the $A_i$. We write $A = x_1 A_1 + ... + x_n A_n$. Thus, we have,

$<x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i>$. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

$0 = x_i <A_i,A_i>$. Because we have presumed that the $A_i ≠ 0$, we cannot have $<A_i,A_i> = 0$, so we must have that the $x_i = 0$. So we have $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0. I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice? 2. Mar 3, 2012 ### Deveno you're fine up to here. at this point, stop thinking about the "inner products", and start thinking about which linear combination of the Ai, A has to be. 3. Mar 3, 2012 ### HallsofIvy You can say more than that- since you are told that the basis is "orthonormal", you know that [itex]<A_i, A_i>= 1$

Once you have $x_i= 0$ for all x, you have immediately that $A= 0A_1+ 0A_2+ ...= 0$

4. Mar 3, 2012

### TranscendArcu

Okay. That makes sense. Thank you both.

This problem is related to the idea of the inner product. I don't think I need any help on the first parts, but I'd like to talk about the third part of the problem.

So, finding the length of a given vector given this inner product:
$<(x,y),(x,y)> = 5x^2 + y^2$.

Taking the length, we have

$|(x,y)| = \sqrt{5x^2 + y^2}$, which we define as equaling 1. Squaring both sides we find,

$5x^2 + y^2 = 1$. I think this is the equation of the circle, but I'm not sure. If it is, then my picture has y-intercepts at 1,-1 and x-intercepts at .5,-.5.

Is this correct?