Question

show that if 0<Ɵ 14 years ago

Answer 1

By definition, if there is an eigenvalue for a linear operator, it can be written as

\[Ax = \lambda x\]
Where (lambda) is the eigenvalue and x is the corresponding eigenvector. Looking at that equation, we are operating the x vector by the transformation A and the result is a multiple of itself (lambda is a constant, so it's equal a constant times itself), meaning that x and Ax are parallel.

That A matrix you wrote over there is called the rotation matrix, which means that if you multiply it by a vector (A*x), the result will be the original vector rotated by an angle (theta) counterclockwise. Therefore the Ax vector will only be parallel to the x vector if you rotate it by: \[\left\{ 0, \pi, 2\pi, 3\pi, … \right\}\]
The interval given in the exercise goes only from 0 to (pi), excluding the endpoints, so the vector transformed by this matrix will never be parallel to the original one => There are no eigenvalues in this interval. I hope it helped.

Answer 2

This answer is fundamentally wrong. It shows that there is no *real* eigenvalue for that matrix, but there are two imaginary eigenvalues, corresponding to the two roots of the characteristic polynomial. These two roots are complex conjugates, and the corresponding eigenvectors are too. For example, if theta = pi/4, then the two eigenvalues are (sqrt(2)/2)(1 + i) and (sqrt(2)/2)(1 - i). These are exp((pi/4)i) and its conjugate exp((-pi/4)i. They are the same "rotation" in the complex plane as theta and -theta.

Answer 3

sorry for answering so late. thank you both very much, it helped a lot.