How do you calculate eigenvalues and eigenvectors of a matrix?

Eigenvalues and eigenvectors are fundamental concepts in linear algebra and have numerous applications in various fields, including mathematical finance. To calculate the eigenvalues and eigenvectors of a square matrix A, you need to follow these steps:

**Step 1: Find the Eigenvalues**

1. First, find the characteristic equation of the matrix. The characteristic equation is given by the determinant of (A − λI), where λ represents the eigenvalue, and I is the identity matrix of the same dimensions as A.

In other words, the characteristic equation is:
det (A − λI) = 0

2. Solve the equation above for the different values of λ. The resulting values are the eigenvalues of the matrix A.

**Step 2: Find the Eigenvectors**

1. For every eigenvalue λ_i, from the first step, find the eigenvector by solving the following equation:
(A − λ_iI)x_i = 0

Note that x_i is the eigenvector corresponding to the eigenvalue λ_i.

2. If the matrix (A − λ_iI) is singular, then it has a non-trivial solution for x_i. The null space of (A − λ_iI) represents the eigenvectors corresponding to the eigenvalue λ_i.

Let’s illustrate these steps with an example.

Consider the following 2x2 matrix:

$$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \\ \end{pmatrix}$$

**Step 1: Find the Eigenvalues**

First, we need to find the determinant of the matrix (A − λI):

$$\det \begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \\ \end{pmatrix} = (2 - \lambda)^2 - 1 \times 1$$

Now, we need to solve the equation (2 − λ)² − 1 = 0:

(2 − λ)² − 1 = λ² − 4λ + 3 = (λ − 1)(λ − 3)

Therefore, the eigenvalues are λ₁ = 1 and λ₂ = 3.

**Step 2: Find the Eigenvectors**

Now, we need to find the eigenvectors corresponding to the eigenvalues we have just found.

1. For λ₁ = 1:

Subtract λ₁I from A and solve the equation for x₁:

$$\begin{pmatrix} 2-1 & 1 \\ 1 & 2-1 \\ \end{pmatrix} \textbf{x}_1 = 0$$

$$\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \textbf{x}_1 = 0$$

The null space of the matrix above corresponds to the unit vector $\textbf{x}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}$.

2. For λ₂ = 3:

Subtract λ₂I from A and solve the equation for x₂:

$$\begin{pmatrix} 2-3 & 1 \\ 1 & 2-3 \\ \end{pmatrix} \textbf{x}_2 = 0$$

$$\begin{pmatrix} -1 & 1 \\ 1 & -1 \\ \end{pmatrix} \textbf{x}_2 = 0$$

The null space of the matrix above corresponds to the unit vector $\textbf{x}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}$.

We have found the eigenvalues and their corresponding eigenvectors:

λ₁ = 1, $\textbf{x}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}$

λ₂ = 3, $\textbf{x}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}$