Diagonalization

Raising a matrix to a power is much easier if the matrix is first diagonalized.

The Essentials

A matrix can be re-written in this form:

\[ A=V^{-1}\Lambda V \]

Where \( \Lambda \) is uppercase lambda and represents a matrix where the diagonal entries are the eigenvalues in some order and all other values are zero, and \( V \) is a matrix whose column vectors are the eigenvectors in the same order as the corresponding eigenvalues. A three by three diagonal matrix has this form:

\[ \begin{bmatrix} * & 0 & 0\\ 0 & * & 0\\ 0 & 0 & * \end{bmatrix} \]

To square A, A can be multiplied by A:

\[ AA=V^{-1} \Lambda VV^{-1} \Lambda V \]
\[ AA=V^{-1} \Lambda I \Lambda V \]
\[ A^2=V^{-1}\Lambda^2 V \]

This can be generalized to this formula:

\[ A^n=V^{-1} \Lambda^n V \]

This is useful because when raising a diagonal matrix to a power, the diagonal entries can just be raised to that power and no matrix multiplication is necessary.

Example

Use diagonalization to evaluate this expression:

\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix}^6 \]

Finding the eigenvalues and eigenvectors, we get:

\[ \lambda_1=-1, x_1= \begin{bmatrix} 2 \\ -1 \end{bmatrix}, \lambda_2=2, x_2= \begin{bmatrix} 1\\ -2 \end{bmatrix} \]

Diagonalizing the matrix we get:

\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix} =-\frac{1}{3} \begin{bmatrix} -2 & -1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0\\ 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1\\ -1 & -2 \end{bmatrix} \]
\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix}^6 =-\frac{1}{3} \begin{bmatrix} -2 & -1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0\\ 0 & 2 \end{bmatrix}^6 \begin{bmatrix} 2 & 1\\ -1 & -2 \end{bmatrix} \]
\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix}^6=-\frac{1}{3} \begin{bmatrix} -2 & -1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 64 \end{bmatrix} \begin{bmatrix} 2 & 1\\ -1 & -2 \end{bmatrix} \]
\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix}^6=-\frac{1}{3} \begin{bmatrix} -2 & -1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1\\ -64 & -128 \end{bmatrix} \]
\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix}^6=-\frac{1}{3} \begin{bmatrix} 60 & 126\\ -126 & -255 \end{bmatrix} \]
\[ \begin{bmatrix} -2 & -2\\ 2 & 3 \end{bmatrix}^6= \begin{bmatrix} -20 & -42\\ 42 & 85 \end{bmatrix} \]

Practice

  1. Diagonalize this matrix:
    \[ \begin{bmatrix} -1 & 1 & 1\\ 0 & 1 & 0\\ 4 & 6 & 2 \end{bmatrix} \]
  2. Evaluate this expression:
    \[ \begin{bmatrix} -2 & -3\\ 1 & 2 \end{bmatrix}^{49} \]

Solutions:

  1. \[ \frac{1}{30} \begin{bmatrix} 0 & -5 & 0\\ -24 & -4 & 6\\ 6 & 21 & 6 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 5 & -1 & 1\\ -6 & 0 & 0\\ 16 & 1 & 4 \end{bmatrix} \]
  2. \[ \begin{bmatrix} -2 & -3\\ 1 & 2 \end{bmatrix} \]