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 &lt;A_i,A_i&gt;. Because we have presumed that the A_i ≠ 0, we cannot have &lt;A_i,A_i&gt; = 0, so we must have that the x_i = 0. So we have &lt;A,A_i&gt; = &lt;x_1 A_1 + ... + x_n A_n,A_i&gt; = 0 = x_1 &lt;A_1,A_i&gt; + ... + x_n &lt;A_n,A_i&gt; = x_i &lt;A_i,A_i&gt; = 0 &lt;A_i,A_i&gt; = &lt;0,A_i&gt; = 0. Which shows that A = 0. <br /> <br /> I don&#039;t know if I&#039;ve done this correctly. I feel like it&#039;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,

&lt;x_1 A_1 + ... + x_n A_n,A_i&gt; = 0 = x_1 &lt;A_1,A_i&gt; + ... + x_n &lt;A_n,A_i&gt;. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

0 = x_i &lt;A_i,A_i&gt;. Because we have presumed that the A_i ≠ 0, we cannot have &lt;A_i,A_i&gt; = 0, so we must have that the x_i = 0.

you're fine up to here.

So we have &lt;A,A_i&gt; = &lt;x_1 A_1 + ... + x_n A_n,A_i&gt; = 0 = x_1 &lt;A_1,A_i&gt; + ... + x_n &lt;A_n,A_i&gt; = x_i &lt;A_i,A_i&gt; = 0 &lt;A_i,A_i&gt; = &lt;0,A_i&gt; = 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,

&lt;x_1 A_1 + ... + x_n A_n,A_i&gt; = 0 = x_1 &lt;A_1,A_i&gt; + ... + x_n &lt;A_n,A_i&gt;. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

0 = x_i &lt;A_i,A_i&gt;. Because we have presumed that the A_i ≠ 0, we cannot have &lt;A_i,A_i&gt; = 0,

You can say more than that- since you are told that the basis is "orthonormal", you know that &lt;A_i, A_i&gt;= 1

so we must have that the x_i = 0. So we have &lt;A,A_i&gt; = &lt;x_1 A_1 + ... + x_n A_n,A_i&gt; = 0 = x_1 &lt;A_1,A_i&gt; + ... + x_n &lt;A_n,A_i&gt; = x_i &lt;A_i,A_i&gt; = 0 &lt;A_i,A_i&gt; = &lt;0,A_i&gt; = 0. Which shows that A = 0. <br /> <br /> I don&#039;t know if I&#039;ve done this correctly. I feel like it&#039;s not entirely convincing towards the end. Advice?

<br /> Once you have x_i= 0 for all x, you have immediately that A= 0A_1+ 0A_2+ ...= 0

 

[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:
&lt;(x,y),(x,y)&gt; = 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?