Matrices, Proof and Eigenvalues.

  • Thread starter chris_avfc
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  • #1
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Homework Statement



Looking for some help with the proof if possible.

Vector r =
x
y
z

Rotation R =
cos(θ) 0 sin(θ)
0 1 0
-sin(θ) 0 cos(θ)

r' = Rr

It asks me to prove that
r'.r' = r.r




Second part of the question is about eigenvalues, it asks me to find the three eigenvalues of R.
I used the formula
det(m - λI)

Where m = R, λ = the eigenvalues and I is the appropriate identity matrix

I end up with λ = cos() or 1, which is clearly wrong as there isn't enough answers.
Somebody in class mentioned imaginary numbers, but I'm unsure as to how to proceed.


The Attempt at a Solution



First Part:

I found r' to be
xcos(θ) + xsinθ
y
-zsin(θ) + xcos(θ)

To get r'.r' am I right to just multiply two of the above together, as in
(xcos(θ) + xsin(θ))(xcos(θ) + xsin(θ))
(yy)
(-zsin(θ) + xcos(θ))(-zsin(θ) + xcos(θ))

Because this is the way I did it and it doesn't lead to the same answer.

Obviously these should all have big brackets around them, but I am unsure of how to represent them on here, if someone would advise me I would gladly fix that.

Second Part:
I used the formula
det(m - λI)

Where m = R, λ = the eigenvalues and I is the appropriate identity matrix

I end up with λ = cos() or 1, which is clearly wrong as there isn't enough answers.
Somebody in class mentioned imaginary numbers, but I'm unsure as to how to proceed.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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964

Homework Statement



Looking for some help with the proof if possible.

Vector r =
x
y
z

Rotation R =
cos(θ) 0 sin(θ)
0 1 0
-sin(θ) 0 cos(θ)

r' = Rr

It asks me to prove that
r'.r' = r.r




Second part of the question is about eigenvalues, it asks me to find the three eigenvalues of R.
I used the formula
det(m - λI)

Where m = R, λ = the eigenvalues and I is the appropriate identity matrix

I end up with λ = cos() or 1, which is clearly wrong as there isn't enough answers.
Somebody in class mentioned imaginary numbers, but I'm unsure as to how to proceed.


The Attempt at a Solution



First Part:

I found r' to be
xcos(θ) + xsinθ
This is incorrect but I expect it is a typo.

y
-zsin(θ) + xcos(θ)

To get r'.r' am I right to just multiply two of the above together, as in
(xcos(θ) + xsin(θ))(xcos(θ) + xsin(θ))
(yy)
(-zsin(θ) + xcos(θ))(-zsin(θ) + xcos(θ))
Because I did it this way and it doesn't lead to the same answer.
Yes, that is what you want to do- and add them. You need to fix that first coordinate of course. What did you get? Don't forget that [itex]cos^2(\theta)+ sin^2(\theta)= 1[/itex].

Obviously these should all have big brackets around them, but I am unsure of how to represent them on here, if someone would advise me I would gladly fix that.

Second Part:
I used the formula
det(m - λI)

Where m = R, λ = the eigenvalues and I is the appropriate identity matrix

I end up with λ = cos() or 1, which is clearly wrong as there isn't enough answers.
Somebody in class mentioned imaginary numbers, but I'm unsure as to how to proceed.
We're not mindreaders! No one can tell you what you did wrong unless you tell us what you did!
 
  • #3
85
0
This is incorrect but I expect it is a typo.



Yes, that is what you want to do- and add them. You need to fix that first coordinate of course. What did you get? Don't forget that [itex]cos^2(\theta)+ sin^2(\theta)= 1[/itex].


We're not mindreaders! No one can tell you what you did wrong unless you tell us what you did!

What's wrong with the first bit, have I made a stupid mistake?

I've attached what I have done.
 

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  • #4
HallsofIvy
Science Advisor
Homework Helper
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You have [itex]det(A-\lambda I)= (cos(\theta)- \lambda)(1- \lambda)(cos(\theta)- \lambda)[/itex] which is wrong. That is the product of the values on the main diagonal but is not the determinant.
 
  • #5
85
0
You have [itex]det(A-\lambda I)= (cos(\theta)- \lambda)(1- \lambda)(cos(\theta)- \lambda)[/itex] which is wrong. That is the product of the values on the main diagonal but is not the determinant.

Ah yeah, I've sorted that all out now.
Thank you so much for the help mate.
 

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