Solving Complex Eigenvalues: Geometric Interpretation

mpm
Messages
82
Reaction score
0
I've got a homework problem that I am needing to do; however, I am not sure really what the question is asking. Obviously since I don't know what is being asked, I don't know where to begin.

I was hoping for some insight.

Question:

Show that matrix

A = {cos (theta) sin (theta), -sin (theta) cos (theta)}

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

Can anyone clear this up for me or help?
 
Physics news on Phys.org
Are you saying that you don't know what is meant by "eigenvalue" or that you just have no idea how to find an eigenvalue?
 
Do you know what complex numbers are? Do you know what eigenvalues of a matrix are? Geometrically, what does that matrix do, i.e. if you drew a line from (0,0) to (x,y) on a cartesian plane, representing the vector (x,y), and then computed A(x,y) to get (a,b), and then drew the line segment from (0,0) to (a,b) on your graph, what will the relationship be between theta, (x,y), and (a,b)? If you don't know the answer, do some actual examples.
 
I know what complex numbers and eigenvalues of matrices are.

I just didnt really know what the question meant by "geometrically show".
 
mpm said:
I just didnt really know what the question meant by "geometrically show".
What does the matrix A do to a given point in (x,y) with a given value of theta?

As AKG suggests, try some examples with different values of theta and different starting points.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top