If a matrix A has distinct eigenvalues (no repeated eigenvalues), then the matrix A can be diagonalized; i.e., there exists a diagonal matrix D and an invertible matrix V such that A = VDV−l. Furthermore, if A = VDV−l where D is a diagonal matrix, then D contains the eigenvalues of the matrix A on its diagonal and the columns of matrix V contain eigenvectors of A associated with the eigenvalues in the corresponding columns of matrix D. For the matrices A in Exercises, compute the eigenvalues and eigenvectors with [V, D] = eig(A), then show that A = VDV−l by comparing A with V*D*inv(V).
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