# Transformations and eigenvalues

1. Oct 27, 2007

### Kudaros

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

Let A be the matrix of the linear transformation T. Without writing A, find an eigenvalue of A and describe the eigenspace. T is the transformation on R^3 that rotates points about some line through the origin.

2. Relevant equations

maybe...Ax=(lambda)x ?

3. The attempt at a solution

The biggest issue for me is imagining this problem geometrically. I am not sure if I understand it.

Any eigenvector will be on this line through the origin that, despite the transformation, remains on that line. So any eigenvalue corresponding to that eigenvector is valid? (and given the abstract question, can I answer it as such? thats also somewhat new and strange to me.)

As for the description of the eigenspace, which is the null space of the matrix A-(lambda)I would that just be a single vector with three entries? I think this because the generation of a line in R^3 is done with the linear combinations of a single vector.

I am imagining that when the question says " rotates points about some line through the origin" the points on the line stay in place, while points outside the line in three space are rotated around this line.

My problem with this class in general, I think, is making the jump from primarily computational problems with some theory mixed in, to primarily theoretical problems, with computational mixed in. I usually enjoy these types of problems, but I think im having an off semester or something.

2. Oct 28, 2007

### javierR

Your reasoning is roughly correct. The operator T acts as rotations in 3-space, but you can think of an arbitrary rotation to consist of a series of rotations, each about a particular line (leaving that line unchanged). The most convenient way to do this is by laying down an xyz coordinate frame, then recreating the arbitrary rotation via a rotation about the x-axis, then the y-axis, then the z-axis; so there are 3 linearly independent types of rotations you can make each one leaving one of the axes unchanged.
Now, an eigenvector x is a vector that appears in the operator equation Tx=lambda x, which you should interpret as: The operator T acts on the vector leaving its direction unchanged but the magnitude changed by a factor of lambda. For rotations this has a simple interpretation, as the eigenvectors can be taken as the *orthonormal basis vectors* of the 3-space (which we coordinatized by x,y and z), so they can be written as three 3-component vectors. Once the 3x3 matrix A is diagonalized for one of the eigenvectors (say, the one for the z-direction) we have
$A_{z}\bf{z}=\lambda_{z}\bf{z}$. (Note that this matrix will not preserve the direction of the other two eigenvectors).
Now, how does the magnitude of the eigenvector z change when you rotate about the z-axis?

3. Oct 28, 2007

### HallsofIvy

Staff Emeritus
This problem is asking you to think "geometrically"
$Ax= \lambda x$ means that A transforms the vector x into a multiple of itself. That is, into a vector lying along the same line as x, with length multiplied by $\lambda$. Do you see that the only vectors that "rotate" into vectors lying along the same line are the vectors lying along the axis of rotation? And what happens to the lengths of vectors under rotation?

4. Oct 28, 2007

### Kudaros

If we are simply rotating the space around the line as described, then the transformation affects its ('its' being the corresponding eigenvector on the line) length by lambda, if at all.

So let x be a vector on this line, and A be the standard matrix for this transformation. Lambda*x is the length of the vector on that line, after the transformation Ax.

Thats how I'm understanding it. Can this line be thought of as the axis of rotation for a body like earth? Does the axis itself rotate?

5. Oct 28, 2007

### HallsofIvy

Staff Emeritus
I was trying to get you to see what lambda is! If you are rotating vectors around a line, what does that do to their lengths?

6. Oct 28, 2007

### Kudaros

Just rotating a vector does nothing to its length, so does that mean lambda is one?

If eigenvalues reflected direction, that would not even matter here would it?

Last edited: Oct 28, 2007