Jan16-11, 09:53 AM
From my Linear Algebra course I learned tha and eigenvalue w is an eigenvalue if it is a sollution to the system:
Ax=wx, where A= square matrix, w= eigenvalue, x= eigenvector. We solved the system by setting det(A-I*w)=0, I=identity matrix
Now in an advanced course I have come upon the equation system Ax=wKx, A= square matrix, x= vector, K= square matrix, w= scalar.
They say we can solve it by setting (A-wK)x=0, and they call this an eigenvalue-problem. And say we can solve it by settig det(A-wK)=0.
My question: Is this really an eigenvalue-problem? I looked in my book and on wikipedia, and there they both say that eigenvalue/vector prblems is Ax=wx, in my problem I have a matrix on both sides. Ax=wKx, so can w then be a eigenvalue?
I appreciate the help.
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