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 λi, from the first step, find the eigenvector by solving the following equation:
(A − λiI)xi = 0
Note that xi is the eigenvector corresponding to the eigenvalue λi.
2. If the matrix (A − λiI) is singular, then it has a non-trivial solution for xi. The null space of (A − λiI) represents the eigenvectors corresponding to the eigenvalue λi.
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, $\textbf{x}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}$