Question

Let \(A\) be a complex (or real) \(n \times n\) matrix, and let \({\bf{x}}\) in \({\mathbb{C}^n}\) be an

eigenvector corresponding to an eigenvalue \(\lambda \) in \(\mathbb{C}\). Show that for each nonzero complex scalar \(\mu \), the vector \(\mu {\bf{x}}\) is an eigenvector of \(A\).

Short answer

It is proved that the vector \(\mu {\bf{x}}\) is an eigenvector of \(A\).

Step-by-step solution

Matrix rules

Twomatricescan be multiplied only when the number of columns in the first equals the number of rows in the second

Calculating the vector

By the definition of eigenvalues and eigenvectors we have \(A{\bf{x}} = \lambda {\bf{x}}\), where \({\bf{x}} \ne {\bf{0}}\) Let \(\mu \ne 0\) be a complex scalar. Then \(\mu {\bf{x}} \ne {\bf{0}}\) and we have;

\(\begin{aligned}{c}A(\mu {\bf{x}}) = \mu (A{\bf{x}})\\ = \mu (\lambda {\bf{x}})\\ = \lambda (\mu {\bf{x}})\end{aligned}\)

The above calculation proves that the vector \(\mu {\bf{x}}\) is an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda \).