21  The Characteristic Equation

library(tidyverse)
library(dasc2594)

The characteristic equation/polynomial encodes information about the eigenvalues of the characteristic equation. In the previous chapter, we showed how we can decide if a scalar \(\lambda\) is an eigenvalue of a matrix and how to find the vectors associated with the eigenvalue. However, we did not learn how to find eigenvalues (other than to just randomly try \(\lambda\)). The characteristic equation/polynomial allows for determining the eigenvalues \(\lambda\).

Definition 21.1 Let \(\mathbf{A}\) be a \(n \times n\) matrix. The characteristic equation/polynomial of \(\mathbf{A}\) is the function \(f(\lambda)\) given by

\[ \begin{aligned} f(\lambda) = det(\mathbf{A} - \lambda \mathbf{I}) \end{aligned} \]

While not obvious, the function \(f(\lambda)\) is a polynomial of \(\lambda\) but requires computing the determinant of the matrix \(\mathbf{A} - \lambda \mathbf{I}\) which contains an unknown value \(\lambda\).

Example 21.1 Find the characteristic equation of the matrix \(\mathbf{A} = \begin{pmatrix} 3 & 5 \\ 2 & -1 \end{pmatrix}\)

  • do in class

Example 21.2 Find the characteristic equation of the matrix \(\mathbf{A} = \begin{pmatrix} 0 & 6 & 8 \\ \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \end{pmatrix}\)

  • do in class (expand cofactors along the third column)

Once the characteristic equation is defined, we can use the equation to solve for the eigenvalues.

Theorem 21.1 Let \(\mathbf{A}\) be a \(n \times n\) matrix and let \(f(\lambda) = det(\mathbf{A} - \lambda \mathbf{I})\) be a characteristic polynomial. Then, the number \(\lambda_0\) is an eigenvalue of \(\mathbf{A}\) if and only if \(f(\lambda_0) = 0\).

By the invertible matrix theorem, the matrix \((\mathbf{A} - \lambda_0 \mathbf{I}) \mathbf{x} = \mathbf{0}\) has a nontrivial solution if and only if \(det(\mathbf{A} - \lambda_0 \mathbf{I}) \mathbf{x} = \mathbf{0}\). Therefore, the following statements are equivalent:

  • \(\lambda_0\) is an eigenvalue of \(\mathbf{A}\)

  • \(\mathbf{A} \mathbf{x} = \lambda_0 \mathbf{x}\) has a nontrivial solution

  • \((\mathbf{A} - \lambda_0 \mathbf{I}) \mathbf{x} = \mathbf{0}\) has a nontrivial solution

  • \(\mathbf{A} - \lambda_0 \mathbf{I}\) is not invertible

  • \(det(\mathbf{A} - \lambda_0 \mathbf{I}) = \mathbf{0}\)

  • \(f(\lambda_0) = 0\)

Example 21.3 Using the characteristic equation of the matrix \(\mathbf{A} = \begin{pmatrix} 3 & 5 \\ 2 & -1 \end{pmatrix}\), solve for the eigenvalues and find a basis for the \(\lambda\)-eigenspaces

  • do in class

Example 21.4 Using the characteristic equation of the matrix \(\mathbf{A} = \begin{pmatrix} 0 & 6 & 8 \\ \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \end{pmatrix}\), solve for the eigenvalues and find a basis for the \(\lambda\)-eigenspaces

  • do in class (expand cofactors along the third column)

21.1 Similarity

The idea behind similar matrices is to understand how the linear transformations implied by the transformation behave. Two matrices are similar if their transformation behavior (rotation, expansion/contraction, etc.) is the same but the coordinates on which the matrix operates are different.

Definition 21.2 The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are said to be similar if there exists an invertible matrix \(\mathbf{P}\) where

\[ \begin{aligned} \mathbf{A} = \mathbf{P} \mathbf{B} \mathbf{P}^{-1} \end{aligned} \]

or equivalently

\[ \begin{aligned} \mathbf{P}^{-1} \mathbf{A} \mathbf{P}= \mathbf{B} \end{aligned} \]

Therefore, it is possible to change \(\mathbf{A}\) into \(\mathbf{B}\) with an invertible (one-to-one and onto) transformation.

Example 21.5 Consider the following example with matrices \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{P}\) defined as below:

A <- matrix(c(3, 0, 0, -2), 2, 2)
A
     [,1] [,2]
[1,]    3    0
[2,]    0   -2
B <- matrix(c(-12, -10, 15, 13), 2, 2)
B
     [,1] [,2]
[1,]  -12   15
[2,]  -10   13
P <- matrix(c(-2, 1, 3,-1), 2, 2)
P
     [,1] [,2]
[1,]   -2    3
[2,]    1   -1
P %*% B %*% solve(P)
     [,1] [,2]
[1,]    3    0
[2,]    0   -2
solve(P) %*% A %*% P
     [,1] [,2]
[1,]  -12   15
[2,]  -10   13

Theorem 21.2 If \(\mathbf{A}\) and \(\mathbf{B}\) are \(n \times n\) similar matrices, then \(\mathbf{A}\) and \(\mathbf{B}\) will have the same characteristic polynomial and therefore the same eigenvalues.

If \(\mathbf{A}\) and \(\mathbf{B}\) are similar, then there exists an invertible matrix \(\mathbf{P}\) such that

\[ \begin{aligned} \mathbf{A} = \mathbf{P} \mathbf{B} \mathbf{P}^{-1} \end{aligned} \]

Therefore

\[ \begin{aligned} \mathbf{A} - \lambda \mathbf{I} & = \mathbf{P} \mathbf{B} \mathbf{P}^{-1} - \lambda \mathbf{I} \\ & = \mathbf{P} \mathbf{B} \mathbf{P}^{-1} - \lambda \mathbf{P} \mathbf{P}^{-1} \\ & = \mathbf{P} \left( \mathbf{B} \mathbf{P}^{-1} - \lambda \mathbf{P}^{-1} \right) \\ & = \mathbf{P} \left( \mathbf{B} - \lambda \mathbf{I} \right) \mathbf{P}^{-1}\\ \end{aligned} \]

To get the characteristic equation, we need to solve for the determinant

\[ \begin{aligned} det\left( \mathbf{A} - \lambda \mathbf{I} \right) & = det\left( \mathbf{P} \left( \mathbf{B} - \lambda \mathbf{I} \right) \mathbf{P}^{-1} \right) \\ & = det\left( \mathbf{P} \right) det\left( \mathbf{B} - \lambda \mathbf{I} \right) det\left(\mathbf{P}^{-1} \right) \\ \end{aligned} \]

We know that \(det\left(\mathbf{P}^{-1} \right)\) = \(\frac{1}{det\left(\mathbf{P} \right)}\) (or, equivalently \(det\left(\mathbf{P} \right) det\left(\mathbf{P}^{-1} \right) = det\left(\mathbf{P} \mathbf{P}^{-1} \right) = det(\mathbf{I}) = 1\)), we have \(det\left( \mathbf{A} - \lambda \mathbf{I} \right) = det\left( \mathbf{B} - \lambda \mathbf{I} \right)\) so that \(\mathbf{A}\) and \(\mathbf{B}\) have the same characteristic polynomial (and the same eigenvalues).

21.2 The geometric interpetation of similar matrices

In general, similar matrices do similar things in different spaces (different spaces in terms of different bases).

Example here