Diagonalize to Power a Matrix: Examples

We begin by trying to diagonalize \(\Mtrx{A}\). The characteristic polynomial of \(\Mtrx{A}\) is

\[ \begin{array}{rcl} p(\lambda) & = & \det \left[ \begin{array}{cc} -9-\lambda & 4 \\ -33 & 14 -\lambda \end{array} \right] \\ & = & -(9+\lambda)(14 - \lambda)\ +\ 132 \\ & = & \lambda^2 - 5\lambda + 6 \\ & = & (\lambda - 2)(\lambda-3) \end{array} \]

So \(\Mtrx{A}\) has two distinct eigenvalues:

1. \(\lambda_1 = 2\) with algebraic multiplicity \(1\)
2. \(\lambda_2 = 3\) with algebraic multiplicity \(1\)

The eigenspace of \(\Mtrx{A}\) associated to \(\lambda_1\) consists of the solutions of the matrix equation

\[ \left[ \begin{array}{rr} -11 & 4 \\ -33 & 12 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right] \]

Therefore, \(E_1 = \span(4,11)\)

The eigenspace of \(\Mtrx{A}\) associated to \(\lambda_2\) consists of the solutions of the matrix equation

\[ \left[ \begin{array}{rr} -12 & 4 \\ -33 & 11 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right] \]

Therefore, \(E_2 = \span(1,3)\)

It follows that \(\EuScript{B}\DefEq ((4,11),(1,3))\) is an ordered basis of \(\RNrSpc{2}\) consisting only of eigenvectors of \(\Mtrx{A}\). Therefore \(\Mtrx{A}\) is diagonalizable, and a diagonalizing matrix is given by

\[ \Mtrx{C} \DefEq \left[ \begin{array}{rr} 4 & 1 \\ 11 & 3 \end{array} \right] \qquad \text{with}\qquad \Mtrx{C}^{-1} = \left[ \begin{array}{rr} 3 & -1 \\ -11 & 4 \end{array} \right] \]

Accordingly,

\[ \Mtrx{D} \DefEq \left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right] = \left[ \begin{array}{rr} 3 & -1 \\ -11 & 4 \end{array} \right] \left[ \begin{array}{rr} -9 & 4 \\ -33 & 14 \end{array} \right] \left[ \begin{array}{rr} 4 & 1 \\ 11 & 3 \end{array} \right] \]

We now have the easy computation of \(\Mtrx{A}^{10}\):

\[ \begin{array}{rcl} \Mtrx{A}^{10} & = & \left(\left[ \begin{array}{rr} 4 & 1 \\ 11 & 3 \end{array} \right] \left[ \begin{array}{rr} 2 & 0 \\ 0 & 3 \end{array} \right] \left[ \begin{array}{rr} 3 & -1 \\ -11 & 4 \end{array} \right] \right)^{10} \\ & = & \left[ \begin{array}{rr} 4 & 1 \\ 11 & 3 \end{array} \right] \left[ \begin{array}{rr} 2 & 0 \\ 0 & 3 \end{array} \right]^{10} \left[ \begin{array}{rr} 3 & -1 \\ -11 & 4 \end{array} \right] \\ & = & \left[ \begin{array}{rr} 4 & 1 \\ 11 & 3 \end{array} \right] \left[ \begin{array}{rr} 2^{10} & 0 \\ 0 & 3^{10} \end{array} \right] \left[ \begin{array}{rr} 3 & -1 \\ -11 & 4 \end{array} \right] \\ & = & \left[ \begin{array}{rr} 3\cdot 2^{12} - 11\cdot 3^{10} & 4\cdot 3^{10} - 2^{12} \\ 33\cdot 2^{10} - 11\cdot 3^{11} & 4\cdot 3^{11} - 11\cdot 2^{10} \end{array} \right] \end{array} \]