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.