Eigenvalues of a Matrix

Eigenvalues (or characteristic values) of a square matrix $ A $ are scalar values $ \lambda $ that satisfy the equation $$ A \mathbf{v} = \lambda \mathbf{v} $$ where $ \mathbf{v} $ is a non-zero vector called an eigenvector (or characteristic vector).

This equation indicates that multiplying the matrix $ A $ by the vector $ \mathbf{v} $ does not change the direction of the vector but only scales its magnitude (or reverses its sign) by the factor $ \lambda $.

To determine the eigenvalues, the characteristic polynomial of the matrix is solved:

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

Here, $ I $ represents the identity matrix with the same dimensions as $ A $.

The solutions $ \lambda $ to this equation are the eigenvalues of the matrix $ A $.

Why are eigenvalues important? Eigenvalues are used in a wide range of applications, such as matrix diagonalization, analyzing linear systems, and more.

A Practical Example

Let’s consider the square matrix $ A $:

$$ A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} $$

To find the eigenvalues, we calculate the characteristic polynomial of $ A $:

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

Where $ I $ is the identity matrix of the same size as $ A $.

$$ A - \lambda I = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 - \lambda & 2 \\ 1 & 3 - \lambda \end{pmatrix} $$

The determinant of this matrix is:

$$ \det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - (2 \cdot 1) $$

$$ \det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 $$

Expanding this expression gives:

$$ \det(A - \lambda I) = 12 - 4\lambda - 3\lambda + \lambda^2 - 2 $$

$$ \det(A - \lambda I) = \lambda^2 - 7\lambda + 10 $$

We then solve the quadratic equation $ \lambda^2 - 7\lambda + 10 = 0 $:

$$ \lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$

Where $ a = 1 $, $ b = -7 $, and $ c = 10 $.

$$ \lambda = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} $$

$$ \lambda = \frac{7 \pm \sqrt{49 - 40}}{2} $$

$$ \lambda = \frac{7 \pm \sqrt{9}}{2} $$

$$ \lambda = \frac{7 \pm 3}{2} $$

$$ \lambda = \begin{cases} \frac{7 - 3}{2} = \frac{4}{2} = 2 \\ \\ \frac{7 + 3}{2} = \frac{10}{2} = 5 \end{cases} $$

Thus, the eigenvalues of the matrix are:

$$ \lambda_1 = 5, \quad \lambda_2 = 2 $$

With the eigenvalues determined, we can calculate the corresponding eigenvectors.

How to Find Eigenvectors?For each eigenvalue, we solve $ (A - \lambda I)\mathbf{v} = 0 $ to find the eigenvectors.

For $ \lambda_1 = 5 $, the matrix $ A - 5I $ is: $$ A - 5I = \begin{pmatrix} 4 - 5 & 2 \\ 1 & 3 - 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} $$ Solving the equation $ (A - 5I)\mathbf{v} = 0 $: $$ \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$ This yields the system: $$ \begin{cases} -x + 2y = 0 \\ \\ x - 2y = 0 \end{cases} $$$$ \begin{cases} -x + 2y = 0 \\ \\ x = 2y \end{cases} $$ Substituting $ x = 2y $: $$ \begin{cases} -(2y) + 2y = 0 \\ \\ x = 2y \end{cases} $$$$ \begin{cases} 0 = 0 \\ \\ x = 2y \end{cases} $$ Since the first equation is always true, the solutions are all vectors satisfying $ x = 2y $. Thus, the eigenvector corresponding to $ \lambda_1 = 5 $ is of the form $ \mathbf{v}_1 = \begin{pmatrix} 2y \\ y \end{pmatrix} $. For example, $ \mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix} $.
For $ \lambda_2 = 2 $, the matrix $ A - 2I $ is: $$ A - 2I = \begin{pmatrix} 4 - 2 & 2 \\ 1 & 3 - 2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} $$ Solving $ (A - 2I)\mathbf{v} = 0 $: $$ \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$ This leads to the system: $$ \begin{cases} 2x + 2y = 0 \\ \\ x + y = 0 \end{cases} $$$$ \begin{cases} 2x + 2y = 0 \\ \\ x = -y \end{cases} $$ Substituting $ x = -y $: $$ \begin{cases} 2(-y) + 2y = 0 \\ \\ x = -y \end{cases} $$$$ \begin{cases} 0 = 0 \\ \\ x = -y \end{cases} $$ The solutions are all vectors satisfying $ x = -y $. Thus, the eigenvector corresponding to $ \lambda_2 = 2 $ is of the form $ \mathbf{v}_2 = \begin{pmatrix} x \\ -x \end{pmatrix} $. For example, $ \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $.

To summarize, the eigenvalues and their corresponding eigenvectors are:

$ \lambda_1 = 5, \quad \mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix} $
$ \lambda_2 = 2, \quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $

Example 2

In this example, we calculate eigenvalues in a system of equations.

Consider the system:

$$ \begin{cases} 3x + 2y &= 6, \\ 4x + y &= 5 \end{cases} $$

Linear systems can be expressed in matrix form as:

$$ A \mathbf{x} = \mathbf{b} $$

Here, $ A $ is the coefficient matrix, $ \mathbf{x} $ is the unknown vector, and $ \mathbf{b} $ is the vector of known terms.

The coefficient matrix is:

$$ A = \begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} $$

Eigenvalues of $ A $ are found by solving the characteristic polynomial:

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

Construct the matrix $ A - \lambda I $:

$$ A - \lambda I = \begin{bmatrix} 3 - \lambda & 2 \\ 4 & 1 - \lambda \end{bmatrix} $$

Then calculate its determinant:

$$ \det(A - \lambda I) = (3 - \lambda)(1 - \lambda) - (4 \cdot 2) $$

$$ \det(A - \lambda I) = (3 - \lambda)(1 - \lambda) - 8 $$

$$ \det(A - \lambda I) = 3 - 3\lambda - \lambda + \lambda^2 - 8 $$

$$ \det(A - \lambda I) = \lambda^2 - 4\lambda - 5 $$

The characteristic polynomial is:

$$ \lambda^2 - 4\lambda - 5 = 0 $$

Solving for the eigenvalues:

$$ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Where $ a = 1 $, $ b = -4 $, and $ c = -5 $.

$$ \lambda = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} $$

$$ \lambda = \frac{4 \pm \sqrt{16 + 20}}{2} $$

$$ \lambda = \frac{4 \pm \sqrt{36}}{2} $$

$$ \lambda = \frac{4 \pm 6}{2} $$

$$ \lambda = \begin{cases} \frac{4 - 6}{2} = \frac{-2}{2} = -1 \\ \\ \frac{4 + 6}{2} = \frac{10}{2} = 5 \end{cases} $$

Therefore, the eigenvalues are $ \lambda_1 = 5 $ and $ \lambda_2 = -1 $.

And so on.