Question

Use mathematical induction to show that if \(\lambda \) is an eigenvalue of an \(n \times n\) matrix \(A\), with a corresponding eigenvector, then, for each positive integer \(m\), \({\lambda ^m}\)is an eigenvalue of \({A^m}\), with \({\rm{x}}\) a corresponding eigenvector.

Short answer

It is proved that \({\lambda ^m}\) is also an eigenvalue of \({A^m}\).

Step-by-step solution

Definition of mathematical induction and Eigen Value of a matrix

Mathematical Inductionis a mathematical technique that is used to prove a

the statement, a formula or a theorem is true for every natural number.

The eigenvalue\(\lambda \) is the real or complex number of a matrix \(A\) which is a square matrix that satisfies the following equation

\(\det \left( {A - \lambda I} \right) = 0\)

This equation is called the characteristic equation.

By using mathematical induction

Proving the by mathematical induction.

For\(n = 1\)

\({A^1}{\rm{x}} = A{\rm{x}} = \lambda {\rm{x}}\)

Suppose that it is true for\(n = m\), hence\({A^m}{\rm{x}} = {\lambda ^m}{\rm{x}}\)for some value\(m > 1\).

Now put \(n = m + 1\), so

\(\begin{array}{c}{A^{m + 1}}{\rm{x}} = {A^1}{A^m}{\rm{x}}\\ = A{\lambda ^m}{\rm{x}}\\ = {\lambda ^m}A{\rm{x}}\\ = {\lambda ^m} \cdot \lambda {\rm{x}}\\ = {\lambda ^{m + 1}}{\rm{x}}\end{array}\)

Hence the statement is true for all \(n\).

Thus \({\lambda ^m}\) is also an eigenvalue of \({A^m}\).