Question

Let A be a complex (or real) n x n matrix, and let x in C^n be an eigenvector corresponding to an eigenvalue (lambda) in C Show that for each nonzero complex scalar (nu) , the vector (nu) x is an eigenvector of A

Answer 1

The vector (nu) x is an eigenvector of A and corresponds to each nonzero complex scalar (nu) (lambda).

Let x in Cn be an eigenvector of A that corresponds to an eigenvalue. Assume that A is a n x n matrix (lambda). Ax = (lambda)x follows.

Let (nu) now be a complex scalar that is nonzero. A((nu)x) = A((nu)x) = A((nu)x) = A((nu)x) = A((nu)x) = A((nu)x)

As a result, (nu)x is an eigenvector of A that corresponds to (nu) (lambda).

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