Is Ax=wKx considered an eigenvalue problem in advanced linear algebra?

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SUMMARY

The equation system Ax=wKx is indeed considered a generalized eigenvalue problem in advanced linear algebra. In this context, A is a square matrix, K is another square matrix, w is a scalar, and x represents the eigenvector. The solution involves setting (A-wK)x=0 and determining the eigenvalues by calculating det(A-wK)=0. This formulation is commonly encountered in differential equations, confirming its relevance in advanced mathematical applications.

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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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w is called a generalised eigenvalue of A and K and x is a generalised eigenvector. This form comes up frequently in differential equations. Click here to read more.
 

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