2011-05-22, 11:17 AM
(This post was last modified: 2011-05-22, 04:55 PM by 2147483647.)
Noah Wrote:Well, if you cannot diagonalizable the matrix, you will not have enough eigenvectors and -values. Therefore you cannot solve through eigenvalues and -vectors, and you need to either represent the problem in another way (probably without matrices), or use some other property to bypass the lack of eigenvectors and -values.
That's disheartening. How, then, do I raise a matrix with a repeated eigenvalue to an arbitrary power without the magical formula? ._.
Noah Wrote:Is there any surprising effect here which surprises you? Of course the identity matrix of any power is the identity matrix.
Ah. It's that earlier I didn't realize that raising a diagonal matrix to a power is equal to raising its terms to the same power, so I was fooling around with the identity matrix in hopes of getting insight as to how to do this.
Noah Wrote:Oh, yeah, forgot to tell about that.
Nothing special, actually. Just work with it as you usually work with imaginary numbers.
The thing is that raising a negative eigenvalue to an arbitrary power returns a complex value. This is strange behavior, since the powers would cause a sort of oscillation of the terms into the complex plane and then back to 0 at the integer powers, meaning that a completely real matrix could become complex in between powers. I'm not sure what this suggests, but this immediately makes me think that a matrix with complex terms can become real at certain non-integer powers. Then it follows that I should find the power that makes all the terms real? How should I go about doing that?
Noah Wrote:I haven't really seen many visual examples of complex matrices. However, the eigenvalues of rotation-matrices are imaginary. In the 2D-plane, you can imagine that it works as magnetism, and that the eigenvectors and eigenvalues are the magnetic field which rotates the plane.
Can you link me to such an image? I'm not sure I understand this, and Google isn't turning up anything.
