Orthogonal Basis and Inner Products

[TranscendArcu]

Homework Statement

The Attempt at a Solution

Since 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 $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0. <br /> <br /> I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice?$

 

[Deveno]

Homework Statement

The Attempt at a Solution

Since 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 $x_i = 0$.

you're fine up to here.

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?

at this point, stop thinking about the "inner products", and start thinking about which linear combination of the A_i, A has to be.

 

[HallsofIvy]

Homework Statement

The Attempt at a Solution

Since 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 $<A_i,A_i> = 0$,

You can say more than that- since you are told that the basis is "orthonormal", you know that $<A_i, A_i>= 1$

so we must have that the $x_i = 0$. So we have $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0. <br /> <br /> I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice?$

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

 

[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?