# Vector Analysis - Similarities on Orthonormal Basis

1. Sep 16, 2013

### math222

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

Let L: R2 → Rn be a linear mapping. We call L a similarity if L stretches all vectors by the same factor. That is, for some δL, independent of v,

|L(v)| = δL * |v|

To check that |L(v)| = δL * |v| for all vectors v in principle involves an infinite number of calculations. Therefore a reduction to a finite number of checks can be useful for applications. Here is one such reduction.

Suppose that for the standard orthonormal basis {e1, e2} of R2, we have that L(e1) and L(e2) are orthogonal and have the same length. Show that L is a similarity. (What is δL?)

2. Relevant equations

Perhaps that orthogonal vectors have dot product zero? I'm not sure otherwise.

3. The attempt at a solution

So far I've been able to show that |L(e1)| = |L(e2)| = δL on the assumption that L is a similarity, but that doesn't really help that much. I'm not sure where to go beyond this.

2. Sep 16, 2013

### Dick

You need to use the orthogonal part. Any vector in R^2 can be written as a*e1+b*e2. What's L(a*e1+b*e2)? Compute its length using the dot product and that L(e1) is orthogonal the L(e2). It's really just the Pythagorean theorem.

3. Sep 16, 2013

### math222

So do I assume that L(a*e1 + b*e2) is in fact similar?

And I'm not sure what you mean by using the dot product and orthogonality to compute the length. is there a formula I'm missing?

4. Sep 16, 2013

### Dick

No, you have PROVE it's similar. The length of a vector is $\sqrt{v \cdot v}$. Is that the one you are missing? Use the linearity of L.

5. Sep 16, 2013

### math222

ahhhh ok i'll try again, thanks for your help.

6. Sep 17, 2013

### math222

ok got it. i basically ended up with a^2 + b^2. thanks again for your help.

7. Sep 17, 2013

### Dick

That's a little sketchy, but it sounds like you doing it right.