Eigenvalues: Matrix corresponding to projection

[shiri]

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.

 

[Dick]

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.

 

[shiri]

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?

 

[Dick]

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?

 

[shiri]

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

 

[Dick]

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.

 

[shiri]

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?

 

[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)?

 

[shiri]

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?

 

[Dick]

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?

 

[shiri]

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?

 

[Dick]

Av is (v.v)v/|v|^2. (v.v) is the same as |v|^2. Av is v!

 

[shiri]

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?

 

[Dick]

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.