Tackling the Characteristic Equation
A Guide to Eigenproblems

Is there a method for calculating all of a matrix’s eigenvalues and corresponding eigenvectors?

Yes!

By finding the roots of the characteristic equation, sometimes called the characteristic polynomial.

Characteristic Equation and Finding Eigenvalues

If \(\mathrm{A}\) is a square \(n \times n\) matrix, then \(\lambda\) is an eigenvalue of matrix \(\mathrm{A}\) if an only if \(\lambda\) satisfies the characteristic equation

\begin{equation}
\operatorname{det}(A-\lambda I)=0
\end{equation}

Example: Calculating Eigenvalues

For example, let’s find the eigenvalues of the following matrix.

\begin{align*}
A=\left[\begin{array}{cc}
4 & -2 \\
-3 & 9
\end{array}\right]
\end{align*}

First, we subtract lambda down the main diagonal to satisfy the expression \(A-\lambda I\)

\begin{aligned}
A-\lambda I & =\left[\begin{array}{cc}
4 & -2 \\
-3 & 9
\end{array}\right]-\lambda\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4 & -2 \\
-3 & 9
\end{array}\right]-\left[\begin{array}{cc}
\lambda & 0 \\
0 & \lambda
\end{array}\right] \\
& =\left[\begin{array}{cc}
4-\lambda & -2 \\
-3 & 9-\lambda
\end{array}\right]
\end{aligned}

Next, we calculate the determinant of this matrix by recalling that \(\operatorname{det}\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)=a d-b c\)

\begin{aligned}
\operatorname{det}\left[\begin{array}{cc}
4-\lambda & -2 \\
-3 & 9-\lambda
\end{array}\right] & =(4-\lambda)(9-\lambda)-(-2)(-3) \\
& =\left(36-13 \lambda+\lambda^{2}\right)-6 \\
& =\lambda^{2}-13 \lambda+30
\end{aligned}

Now we solve for the roots of the characteristic equation by setting the polynomial equal to zero and solving.

\begin{aligned}
\lambda^{2}-13 \lambda+30 & =0 \\
(\lambda-10)(\lambda-3) & =0 \\
\lambda & =10 \text { or } \lambda=3
\end{aligned}

So, all we have to do to find eigenvalues is solve the characteristic A polynomial.

Calculating Eigenvectors

Okay, but how do we find each subsequent eigenvector, which will yield the basis for the eigenspace? We find the \(\operatorname{Nul}(A-\lambda I)\) by substituting our eigenvalues into our matrix equation and solving the homogeneous equation.

Example: Calculating Eigenvectors

Let’s continue our example from above, as we have already found our two eigenvalues and used them to find the eigenspace.

Find a basis for the eigenspace corresponding to
\[
A=\left[\begin{array}{cc}
4 & -2 \\
-3 & 9
\end{array}\right]
\]
if \(\lambda=10\) and \(\lambda=3\).

We will work with each eigenvalue separately.

So, if \(\lambda=10\), then
\begin{aligned}
A-\lambda I
& = \left[\begin{array}{cc}
4-10 & -2 \\
-3 & 9-10
\end{array}\right] \\
& = \left[\begin{array}{cc}
-6 & -2 \\
-3 & -1
\end{array}\right]
\end{aligned}

Now we augment with the zero vector and row reduce:

\begin{aligned}
\left[\begin{array}{ll}
(A-\lambda I) & \overrightarrow{0}
\end{array}\right]
& = \left[\begin{array}{lll}
-6 & -2 & 0 \\
-3 & -1 & 0
\end{array}\right] \\
& \sim \left[\begin{array}{ccc}
1 & 1 / 3 & 0 \\
0 & 0 & 0
\end{array}\right] \\
\begin{array}{l}
x_1 = -\frac{1}{3} x_2 \\
x_2 = x_2
\end{array}
\end{aligned}

So, the eigenvector for \(\lambda=10\) is the null space of the matrix equation written in parametric form.

\begin{aligned}
\lambda=10
& \rightarrow \left[\begin{array}{c}
-1 / 3 \\
1
\end{array}\right] \text { or } \left[\begin{array}{c}
-1 \\
3
\end{array}\right]
\end{aligned}

Next, we will find the eigenvector for the eigenvalue \(\lambda=3\) using the same steps.

\begin{aligned}
A-\lambda I
& = \left[\begin{array}{cc}
4-3 & -2 \\
-3 & 9-3
\end{array}\right] \\
& = \left[\begin{array}{cc}
1 & -2 \\
-3 & 6
\end{array}\right] \\
\left[\begin{array}{cc}
(A-\lambda I) & \overrightarrow{0}
\end{array}\right]
& = \left[\begin{array}{ccc}
1 & -2 & 0 \\
-3 & 6 & 0
\end{array}\right] \\
& \sim \left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& \quad \begin{array}{l}
x_{1}=2 x_{2} \\
x_{2}=x_{2}
\end{array} \\
\lambda=3
& \rightarrow \left[\begin{array}{l}
2 \\
1
\end{array}\right]
\end{aligned}

Eigenspace and Its Basis

Therefore, the basis for the eigenspace is the collection of all eigenvectors for matrix \(\mathrm{A}\), which is

\begin{align*}
\left\{\left[\begin{array}{c}
-1 \\
3
\end{array}\right],\left[\begin{array}{l}
2 \\
1
\end{array}\right]\right\}
\end{align*}

Awesome!

Next Steps

In this video, you will:

• Learn how to construct the characteristic polynomial and solve for eigenvalues
• Utilize your knowledge of homogeneous linear equations to solve for the basis of each eigenspace (i.e., find the eigenvectors)
• Discuss important properties involving the Invertible Matrix Theorem (IMT), the characteristic polynomial, triangular matrices, and similarity
• Explore how to find Eigenvalues and Eigenvectors as they apply to Markov Chains

Get ready for an engaging experience, and let’s get started!

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