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Vector Analysis - Similarities on Orthonormal Basis

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  • #1
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Homework Statement



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?)


Homework Equations



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

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.
 

Answers and Replies

  • #2
Dick
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Homework Statement



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?)


Homework Equations



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

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.
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
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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
Dick
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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?
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
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ahhhh ok i'll try again, thanks for your help.
 
  • #6
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ok got it. i basically ended up with a^2 + b^2. thanks again for your help.
 
  • #7
Dick
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ok got it. i basically ended up with a^2 + b^2. thanks again for your help.
That's a little sketchy, but it sounds like you doing it right.
 

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