Vector Analysis - Similarities on Orthonormal Basis

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Homework Help Overview

The discussion revolves around the concept of linear mappings in vector spaces, specifically focusing on the definition and properties of similarities in the context of a linear mapping L: R2 → Rn. The original poster presents a problem that requires demonstrating that L is a similarity based on the behavior of L applied to the standard orthonormal basis vectors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of L being a similarity and the conditions under which this holds true. There is discussion about the orthogonality of the images of the basis vectors and the use of the dot product to compute lengths. Some participants question how to apply these concepts to the general case of a vector in R2.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to approach the problem. There is a recognition of the need to prove that L(a*e1 + b*e2) is similar, and some participants have made progress in their reasoning but have not yet reached a consensus on the next steps.

Contextual Notes

Participants note the importance of using the properties of orthogonal vectors and the dot product in their calculations. There is also mention of the need to reduce the infinite checks required for all vectors to a finite number based on the orthonormal basis.

math222
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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.
 
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math222 said:

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.
 
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?
 
math222 said:
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.
 
ahhhh ok i'll try again, thanks for your help.
 
ok got it. i basically ended up with a^2 + b^2. thanks again for your help.
 
math222 said:
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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