Do Rotational Matrices Always Yield Real Eigenvalues?

[3.141592654]

Homework Statement

Show that the matrix

A = [cos θ -sin θ
sin θ cos θ]

will have complex eigenvalues if θ is not a multiple of π. Give a geometric interpretation of this result.

Homework Equations

Ax = λx, so

det(A-λI) = 0

The Attempt at a Solution

In this case,

A-λI = [(cos θ - λ) -sin θ
sin θ (cos θ - λ)]

so calculating the determinate, I have:

$det(A-λI) = (cos θ - λ)(cos θ - λ) + sin^{2} θ$

$= cos^{2} θ - 2λcos θ + λ^{2} + sin^{2} θ$

$= λ^{2} - 2λcos θ + 1$

Setting this polynomial to zero, I have

$p(λ) = λ^{2} - 2λcos θ + 1 = 0$

When cos θ is not a multiple of π, I need to use quadratic formula to solve this equation. Doing so, I get:

$λ = cos θ \pm sin θ$

However, this value is not complex, is it? I'm a bit confused on what's going on.

 

[Dick]

You slipped up a bit with the quadratic equation. You don't just get sin(theta). Check it and show what you did if you still don't see it.

 

[3.141592654]

Ah, thank you, after fixing the algebra I see the answer is:

$λ = cos θ ± isin θ.$

In terms of what this means geometrically, it's been a long time since I've done anything with complex numbers, so can you correct the following if it is wrong?

I know that since λ is complex, the corresponding eigenvectors must also be complex, so θ is the argument of the eigenvectors in the complex plane. In that case, the matrix A (and hence λ) multiplies each element, z = x + iy, of the vector x, by cos ± isin θ.

When I multiply λ*z, I get:

$λz = (xcos θ + ysin θ) + i(cos θ sin θ y)$

So the matrix A essentially makes the real component of each element of x a combination of x and y, where xcos θ + ysin θ ≤ x + y, and it multiplies the imaginary component by cos θ sin θ. In conjunction, these effects would shrink the imaginary component of z and could shrink or stretch the real component.

Is this correct? Is there something else that the problem is hoping that I'll recognize about how lambda acts on the vector x?

Thanks.

 

[3.141592654]

I've realized that the matrix given in this problem is an orthogonal matrix, so it rotates the vector x. However, should I be able to see that lamda (and A) perform a rotation if I didn't realize that A was an orthogonal matrix?

 

[Dick]

I've realized that the matrix given in this problem is an orthogonal matrix, so it rotates the vector x. However, should I be able to see that lamda (and A) perform a rotation if I didn't realize that A was an orthogonal matrix?

It's a matrix you've probably seen a lot of times in lower level courses. I think they might have expected you to recognize it, even if they didn't use the words 'orthogonal' back then. It rotates a vector in the x-y plane by an angle theta counter-clockwise. You can do the same thing regarding a point in the x-y plane as a complex number (the Argand plane) and multiplying by cos(theta)+i*sin(theta)=cis(theta)=exp(i*theta). I'm having trouble coming up with anything more formal than that just now.

 

[Office_Shredder]

Geometric interpretation: Which rotations of the plane have eigenvectors? For example, if I rotate the point (1,0) and get (x,0) for some value of x, what were the possible rotations that could have given me that