# Matrices, Proof and Eigenvalues.

1. Mar 18, 2012

### chris_avfc

1. The problem statement, all variables and given/known data

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.

3. 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.

2. Mar 18, 2012

### HallsofIvy

Staff Emeritus
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 $cos^2(\theta)+ sin^2(\theta)= 1$.

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

3. Mar 18, 2012

### chris_avfc

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. Mar 18, 2012

### HallsofIvy

Staff Emeritus
You have $det(A-\lambda I)= (cos(\theta)- \lambda)(1- \lambda)(cos(\theta)- \lambda)$ which is wrong. That is the product of the values on the main diagonal but is not the determinant.

5. Mar 19, 2012

### chris_avfc

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