What is a Characteristic Polynomial of a Matrix? – Fusion of Engineering, Control, Coding, Machine Learning, and Science

In this mathematics, matrix algebra, and control engineering tutorial, we explain the concept of the characteristic polynomial. The characteristic polynomial is used to compute the eigenvalue of a matrix and in control engineering, it is used to verify the stability of linear dynamical systems. The YouTube tutorial accompanying this webpage is given below.

Let us start with the motivation and background. Let us assume that we are interested in finding eigenvalues and eigenvectors of a matrix $A$ . Let the eigenvalue be denoted by $\lambda$ and let the eigenvector be denoted by $\mathbf{w}$ . Then, the eigenvector and eigenvalue satisfy this equation

(1) $\begin{align*}A\mathbf{w}=\lambda \mathbf{w}\end{align*}$

Loosely speaking, this equation tells us that if we multiply the matrix $A$ by its eigenvector $\mathbf{w}$ , then the result will be the vector $\lambda \mathbf{w}$ that has the same action line as the original eigenvector $\mathbf{w}$ . If the eigenvalue $\lambda$ is real, then the vector $\lambda \mathbf{w}$ is a scaled version (with a possibility of the direction change) of the vector $\mathbf{w}$ .

The main question is how to find the eigenvalues and eigenvectors from (1). Let us do that. From (1), we have

(2) $\begin{align*}& A\mathbf{w}-\lambda \mathbf{w}=0 \\& \big( A-\lambda \cdot I \big) \mathbf{w}=0\end{align*}$

where $I$ is an identity matrix (matrix with all the entries on the main diagonal equal to 1). This is a system of equations in the unknown vector $\mathbf{w}$ . This system obviously has a trivial solution $\mathbf{w}=0$ (a trivial solution is a vector whose all entries are equal to zero). However, we are not interested in a trivial solution. We are interested in a nontrivial solution (a nontrivial solution is a vector whose entries are not all equal to zero at the same time. That is, at least one entry is not equal to zero). The nontrivial solution exists if and only if

(3) $\begin{align*}\text{det}\Big[\big( A-\lambda \cdot I \big)\Big]=0\end{align*}$

where $\text{det}\Big[\cdot \Big]$ is the matrix determinant. The equation (3) is called the characteristic equation of the matrix $A$ and the polynomial $p(\lambda)=\text{det}\Big[\big( A-\lambda \cdot I \big)\Big]$ is called the characteristic polynomial of the matrix $A$ . This polynomial is the polynomial whose independent variable is $\lambda$ .

Let us now formally define the characteristic equation and the characteristic polynomial.

Definition of the characteristic equation and the characteristic polynomial. Consider the matrix $A\in \mathbb{R}^{n\times n}$ .
– The characteristic polynomial $p(\lambda)$ of the matrix $A$ is defined by

(4) $\begin{align*}p(\lambda)=\text{det}\Big[\big( A-\lambda \cdot I \big)\Big]\end{align*}$

– The characteristic equation of the matrix $A$ is defined by

(5) $\begin{align*}p(\lambda)=\text{det}\Big[\big( A-\lambda \cdot I \big)\Big]=0\end{align*}$

The solutions of the characteristic equation (5) are the eigenvalues of the matrix $A$ . Equivalently, the roots of the characteristic polynomial (4) are the eigenvalues of the matrix $A$ .

Finally, let us do an example to illustrate the computation of the characteristic polynomial and the characteristic equation.

Problem: Compute the characteristic polynomial and eigenvalues of the matrix $A$ :

(6) $\begin{align*}A=\begin{bmatrix} 7 & -4 \\ 5 & -2 \end{bmatrix}\end{align*}$

To solve this problem, first we need to form the matrix $A-\lambda \cdot I$ . We have

(7) $\begin{align*}A-\lambda \cdot I =\begin{bmatrix} 7 & -4 \\ 5 & -2 \end{bmatrix} -\lambda \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 -\lambda & -4 \\ 5 & -2-\lambda \end{bmatrix}\end{align*}$

Next, we need to compute the characteristic polynomial

(8) $\begin{align*}p(\lambda)=\text{det}\Big( \big( A-\lambda \cdot I \big)\Big)=\text{det}\Big( \begin{bmatrix} 7 -\lambda & -4 \\ 5 & -2-\lambda \end{bmatrix} \Big)=\lambda^{2}-5\lambda +6=(\lambda - 2)(\lambda - 3)\end{align*}$

From the characteristic polynomial, we can see that the eigenvalues are $\lambda =2$ and $\lambda =3$ .

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