Matrix - Characteristic Equation

The characteristic equation of a square matrix $A$ is given by the equation:

$$\det (A-\lambda I)=0$$

where:

• $A$ is the square matrix.
• $\lambda$ represents the eigenvalues of the matrix $A$.
• $I$ is the identity matrix of the same size as $A$.

If $A\in n\times n$, then $\det (A-\lambda I)$ is a polynomial of degree $n$ in terms of $\lambda$. As an eigenvalue can be complex, we know there must be at least one eigenvalue for any square matrix.

Example 1

Find the eigenvalue of the matrix:

$$A=\left[\begin{array}{cc}2& 3\\ 3& -6\end{array}\right]$$

using the characteristic equation

$$\det (A-\lambda I)=0$$

Solution:

First, we compute $A-\lambda I$:

$$A-\lambda I=\left[\begin{array}{cc}2-\lambda & 3\\ 3& -6-\lambda \end{array}\right]$$

Next, we compute the determinant:

$$\det (A-\lambda I)=(2-\lambda )(-6-\lambda )-(3)(3)$$

Expanding this, we get:

$$\det (A-\lambda I)={\lambda }^2+4\lambda -21$$

Setting the determinant equal to zero gives us the characteristic equation:

$${\lambda }^2+4\lambda -21=0$$

Factoring the quadratic, we find:

$$(\lambda +7)(\lambda -3)=0$$

Thus, the eigenvalues are:

$${\lambda }_1=-7,{\lambda }_2=3$$

Both these factors have a multiplicity of $1$ each. This means that each eigenvalue corresponds to an eigenspace of dimension $1$.

Example 2

Consider the matrix and eigenvalue:

$$A=\left[\begin{array}{cc}9& 3\\ 3& 1\end{array}\right],\lambda -7$$

Then applying this to the eigenvector equation,

$$(A-\lambda I)\vec{v}=0\Rightarrow (A+7I)\vec{v}=0$$

we get:

$$\left[\begin{array}{cc}9& 3\\ 3& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=0$$

once again, constructing an augmented matrix and solving for the possible values of $\vec{x}$, we get:

$$\left[\begin{array}{ccc}1& \frac{1}{3}& 0\\ 0& 0& 0\end{array}\right]$$

Let $x_2=s$ for some $s\in \mathbb{R}$, then we have:

$$\vec{v}=\left[\begin{array}{c}-\frac{1}{3}s\\ s\end{array}\right]=s\left[\begin{array}{c}-\frac{1}{3}\\ 1\end{array}\right]$$

Therefore, $\left\{\left[\begin{array}{c}-\frac{1}{3}\\ 1\end{array}\right]\right\}$ is a basis for the eigenspace corresponding to the eigenvalue $\lambda =-7$.