What are the complex eigenvalues and eigenvectors of a 2x2 rotation matrix?

[spaghetti3451]

Homework Statement

The 2x2 matrix representing a rotation of the xy-plane is T =

cos θ -sin θ
sin θ cos θ

Show that (except for certain special angles - what are they?) this matrix has no real eigenvalues. (This reflects the geometrical fact that no vector in the plane is carried into itself under such a rotation; contrast rotations in three dimensions.) This matrix does, however, have complex eigenvalues and eigenvectors. Find them. Construct a matrix S which diagonalises T. Perform the similarity transformation (STS^-1) explicitly, and show hat it reduces T to diagonal form.

Homework Equations

The Attempt at a Solution

The characteristic equation is $[cos(θ) - \lambda]^{2}+ [sin^{2}θ] = 0$, that is,
$\lambda^{2} - 2 \lambda cos(θ) + 1 = 0$. The discriminant for this equation is 2i sin(θ). This means that the matrix has real eigenvalues iff θ = nπ for all integers n.


The geometrical interpretation is that in the xy-plane, any arbitrary vector does not get mapped onto its original direction unless there is no rotation (θ = 0), the rotation is half a revolution (θ = ±180°), the rotation is a full revolution (θ = ±360°), ... . The vector (0,0) has a special status in this transformation as it gets mapped onto itself, however, by definition, (0,0) is not an eigenvector.

In three dimensions, again, any arbitrary vector does not get mapped onto its original direction unless there is no rotation (θ = 0), the rotation is half a revolution (θ = ±180°), the rotation is a full revolution (θ = ±360°), ... . The rotation is about a specified axis passing through the origin (the axis has to pass through the origin, otherwise the transformaion is not linear), and all the vectors that live in the axis have a special status in this transformation as they get mapped onto themselves, however, by definition, (0,0) is not an eigenvector, so all the vectors in the axis except for (0,0) are eigenvectors. This is in contrast to the vectors in a plane as such vectors do not exist in that case.



The eigenvalues are $\lambda = e^{±iθ}$. Substitute $\lambda = e^{iθ}$ and eigenvector = {x,y} into the eigenvalue equation to obtain two equations which can be solved to obtain y = -ix. Then, substitute $\lambda = e^{-iθ}$ and eigenvector = {x,y} into the eigenvalue equation to obtain two equations which can be solved to obtain y = ix. So, the eigenvalues are $e^{iθ}$ and $e^{-iθ}$ and the corresponding eigenvectors are {1, -i} and {1, i}.



The matrix S^-1 is given by
1 1
-i i
From this, the matrix S is found to be
1/2 i/2
1/2 -i/2

Performing the similarity transformation (STS-1) explicitly results in the matrix
$e^{iθ}$ 0
0 $e^{-iθ}$.



I would greatly appreciate any comments on the solution.

 

[tiny-tim]

hi failexam!

yes that's fine , except …

i] in 3D, there's an extra row and column, with a 1 and 0s, so there's an extra eigenvalue of 1 (and the question only asks about a plane anyway )

ii] you could divide by detS, so that both S and S^-1 have 1/√2s, which would make the S and S^-1 pair look more symmetric (but you don't have to, since the question doesn't ask for it)