Eigenvalues: Matrix corresponding to projection

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Homework Help Overview

The discussion revolves around the properties of eigenvalues and eigenvectors related to a matrix that represents a projection in two dimensions onto a line defined by a vector v. Participants are analyzing various statements regarding the eigenvalues associated with this projection matrix.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are examining the validity of specific eigenvalue claims for the projection matrix, questioning the necessity of non-zero eigenvalues, and discussing the implications of eigenvectors associated with different eigenvalues. There is also exploration of the relationship between the projection operation and the eigenvalues.

Discussion Status

The discussion is ongoing, with participants expressing differing opinions on the truth of various statements regarding eigenvalues and eigenvectors. Some participants have provided reasoning for their beliefs, while others have challenged these interpretations, leading to a deeper examination of the concepts involved.

Contextual Notes

There is a focus on the definitions and properties of projection matrices, eigenvalues, and eigenvectors, with references to textbook definitions and specific mathematical properties. Some participants are also reflecting on the implications of linear independence in the context of eigenvectors.

shiri
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Let A be a matrix corresponding to projection in 2 dimensions onto the line generated by a vector v.

A) lambda = −1 is an eigenvalue for A
B) The vector v is an eigenvector for A corresponding to the eigenvalue lambda = −1.
C) lambda = 0 is an eigenvalue for A
D) Any vector w perpendicular to v is an eigenvector for A corresponding to the eigenvalue lambda = −1.
E) Any vector w perpendicular to v is an eigenvector for A corresponding to the eigenvalue lambda = 0.

A) A matrix must have a non-zero eigenvalue.
B) The vector is an eigenvector of corresponding to the eigenvalue.
C) See A
D) When a vector is perpendicular to A, it must be zero for eigenvalue (If it was nonzero eigenvalue, then it is not perpendicular)
E) See D

So far I believe A, B, E are true? What do you think? Please help me.
 
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Thanks for giving reasons this time, but I don't believe A) is true. And I believe your reason even less. Let's just start from there. A matrix doesn't HAVE to have a nonzero eigenvalue, much less (-1). I will say your answers to D and E seem to be correct, but if you think E is true, then why do you think C is false? That's just crazy.
 
Dick said:
Thanks for giving reasons this time, but I don't believe A) is true. And I believe your reason even less. Let's just start from there. A matrix doesn't HAVE to have a nonzero eigenvalue, much less (-1). I will say your answers to D and E seem to be correct, but if you think E is true, then why do you think C is false? That's just crazy.

When I read the textbook, it stated that "a square matrix A is invertible if and only if lambda = 0 is not an eigenvalue of A."

Therefore, I assume C) was false.

After I read your message, it appears that B, C, E are correct for this problem?
 
Who said A was invertible? I'm not going to tell you what's correct until you convince me you know what's correct. Why do you think B is true?
 
Dick said:
Who said A was invertible? I'm not going to tell you what's correct until you convince me you know what's correct. Why do you think B is true?

I would say that eigenvector corresponding to the eigenvalue λ = -1 is any multiple of the basic vector. So, both make the eigenspace corresponding to the eigenvalue.

or

The eigenvector corresponding to the eigenvalue λ = −1 is the solution of the equation Ax = -x
 
If you project v onto the line generated by the vector v (which is what A does to v), what do you get? They aren't talking about just any vector that happens to have eigenvalue -1. They are talking about that particular v.
 
Last edited:
Dick said:
If you project v onto the line generated by the vector v (which is what A does to v), what do you get? They aren't talking about just any vector that happens to have eigenvalue -1. They are talking about that particular v.

I would get linearly indepedent eigenvectors, Dick?
 
Do you know what a projection is? http://www.freewebs.com/xinyeeisme/ProjectionVectors_1000.gif That picture shows you the projections of some vectors onto the line generated by the vector w. What's proj_w(w)?
 
Dick said:
Do you know what a projection is? http://www.freewebs.com/xinyeeisme/ProjectionVectors_1000.gif That picture shows you the projections of some vectors onto the line generated by the vector w. What's proj_w(w)?

proj_w(w) = ((w (dot product) w)/w)/||w||^2, isn't?
 
  • #10
shiri said:
proj_w(w) = ((w (dot product) w)/w)/||w||^2, isn't?

Right. So what is Av if A is the projection onto the line generated by v?
 
  • #11
Dick said:
Right. So what is Av if A is the projection onto the line generated by v?

It will be parallel? so, there are many vectors that could correspond to the given?

That means C and E are true, right?
 
Last edited:
  • #12
Av is (v.v)v/|v|^2. (v.v) is the same as |v|^2. Av is v!
 
  • #13
Dick said:
Av is (v.v)v/|v|^2. (v.v) is the same as |v|^2. Av is v!

Therefore C and E are true, right?
 
  • #14
shiri said:
Therefore C and E are true, right?

Av=v has nothing to do with whether C and E are true or false. It's about a different part of the problem.
 

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