Finding eigenvalues For which  is there a non-zero solution to Ax = x Ax = x  Ax – x = 0  Ax – Ix = 0  (AI) x = 0.
{"Image DocumentName": "Finding eigenvaluesFor which \uf06c is there a non-zero solution to Ax = \uf06cx ? Ax = \uf06cx \uf0db Ax \u2013 \uf06cx = 0 \uf0db Ax \u2013 \uf06cIx = 0 \uf0db (A\uf02d\uf06cI) x = 0.", "Image ImageDescription": "So, non trivial solution exists \uf0db det(A \u2013\uf06c I) = 0. \uf044A(\uf06c) = det(A \u2013\uf06c I) is a polynomial of degree n. It is called the characteristic polynomial of A. The roots of \uf044A are the eigenvalues of A. Therefore, there are always at least complex eigenvalues. If n is odd, there is at least one real eigenvalue.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}