FIRST ENCOUNTER WITH EIGENVALUES AND EIGENVECTORS

Consider the matrix $A = \begin{bmatrix}2 & -7 \\ -2 & -3 \end{bmatrix}$ , which can be viewed as a matrix representation of a linear operator on $\mathbb{R}^2$ , and the vector $\vec{x} = \begin{bmatrix}7 \\ -2 \end{bmatrix} \in \mathbb{R}^2$ . It is easy to check that $A \vec{x} = 4 \vec{x}$ . In this example, the output of the matrix transformation is a scalar multiple of the nonzero input vector. Similarly, the nonzero vector $\vec{y} = \begin{bmatrix}1 \\ 1 \end{bmatrix}$ satisfies $A \vec{y} = -5 \vec{y}$ . We can say that 4 is an eigenvalue of A and $\vec{x}$ is an eigenvector corresponding to the eigenvalue 4. Likewise, we can say that -5 is an eigenvalue of A and $\vec{y}$ is an eigenvector corresponding to the eigenvalue -5.

Definition

If A is an n×n matrix, then a nonzero vector $\vec{x}$ in $\mathbb{R}^n$ is called an eigenvector of A if $A \vec{x}$ is a scalar multiple of $\vec{x}$ , that is:

$A \vec{x} = \lambda \vec{x}$ ………………………………………………………………………………………………………………………………………………………………………………………………. (*)

for some scalar λ. The scalar λ is called an eigenvalue of A and $\vec{x}$ is said to be an eigenvector corresponding to λ.

In the definition above, (*) can be written as $A \vec{x} = \lambda I \vec{x}$ where $I$ is the n×n identity matrix. Furthermore, (*) is equivalent to:

$(A - \lambda I) \vec{x} = \vec{0}$ ………………………………………………………………………………………………………………………………………………………………………………… (**)

By the definition above, the eigenvector is not a zero vector. Thus, (**) must have nontrivial solutions. In order that (**) has nontrivial solutions, the determinant of $(A - \lambda I)$ must be zero. Then we have the following equation.

$\vert A - \lambda I \vert = 0$ ……………………………………………………………………………………………………………………………………………………………………………………. (***)

This is called the characteristic equation of A. When expanded, the determinant $\vert A - \lambda I \vert$ is a polynomial in λ and it is called the characteristic polynomial of A.

Example 1

Find the eigenvalues of the matrix $A = \begin{bmatrix}2 & -7 \\ -2 & -3 \end{bmatrix}$ .

Answer

$A - \lambda I = \begin{bmatrix}2 & -7 \\ -2 & -3 \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}2 - \lambda & -7 \\ -2 & -3 - \lambda \end{bmatrix}$

The characteristic polynomial of A is $\left| \begin{matrix}2 - \lambda & -7 \\ -2 & -3 - \lambda \end{matrix} \right| = \lambda^2 + \lambda - 20$ . Hence the characteristic equation of A is λ² + λ – 20 = 0. Solving this quadratic equation, we get two distinct roots, that is λ₁ = 4 and λ₂ = -5.

The task is now to find the corresponding eigenvectors corresponding to the eigenvalues of a given matrix. The eigenvectors of A corresponding to an eigenvalue λ are the nonzero vectors $\vec{x}$ that satisfy (*) above. This is to say that the eigenvectors corresponding to λ are the nonzero vectors in the solution space of $(A - \lambda I) \vec{x} = \vec{0}$ . The solution space is called the eigenspace of A corresponding to λ.

Theorem

If A is an n×n matrix, then the following are equivalent.

λ is an eigenvalue of A.
The system of equations $(A - \lambda I) \vec{x} = \vec{0}$ has nontrivial solutions.
There is a nonzero vector $\vec{x} \in \mathbb{R}^n$ such that $A \vec{x} = \lambda \vec{x}$ .
λ is a real solution of the characteristic equation $\vert A - \lambda I \vert = 0$ .

The example below shows how to apply the theorem to determine the eigenspace of a square matrix.

Example 2

Referring to Example 1, find the eigenspace of A.

Answer

By definition, $\vec{x} = \begin{bmatrix}x_1 \\ x_2 \end{bmatrix}$ is an eigenvector of A corresponding to λ if and only if $\vec{x}$ is a nontrivial solution of $(A - \lambda I) \vec{x} = \vec{0}$ . Thus in this case we have to find nontrivial solutions of the following equation.

$\begin{bmatrix}2 - \lambda & -7 \\ -2 & -3 - \lambda \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$ ………………………………………………………………………………………………………………………………………………… (1)

Example 1 shows that the eigenvalues of A are λ₁ = 4 and λ₂ = -5.

If λ = λ₁ = 4, then (1) becomes:

$\begin{bmatrix}-2 & -7 \\ -2 & -7 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$

which corresponds to the linear equation -2x₁ – 7x₂ = 0. It has infinitely many solutions. It is easy to check that for every $t \in \mathbb{R}$ the ordered pair (7t,-2t) is a solution of the equation. Therefore, the eigenspace of A corresponding to λ = 4 is $E_1 = \begin{Bmatrix} t \begin{bmatrix} 7 \\ -2 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ .

If λ = λ₂ = -5, then (1) becomes:

$\begin{bmatrix}7 & -7 \\ -2 & 2 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$

By applying some elementary row operations, it can be proved that the matrix $\begin{bmatrix}7 & -7 \\ -2 & 2 \end{bmatrix}$ is equivalent to $\begin{bmatrix}1 & -1 \\ 0 & 0 \end{bmatrix}$ . This corresponds to the linear equation x₁ – x₂ = 0. It has infinitely many solutions. It is easy to check that for every $t \in \mathbb{R}$ the ordered pair (t,t) is a solution of the equation. Therefore, the eigenspace of A corresponding to λ = -5 is $E_2 = \begin{Bmatrix} t \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ .

Example 3

Consider the matrix: $A = \begin{bmatrix}5 & 6 & 2 \\ 0 & -1 & -8 \\ 1 & 0 & -2 \end{bmatrix}$ . Find the eigenspace.

Answer

The theorem above suggests that to get the eigenvalues of A we have to find the solution of the characteristic equation $\vert A - \lambda I \vert = 0$ .

$\left| \begin{matrix}5 - \lambda & 6 & 2 \\ 0 & -1 - \lambda & -8 \\ 1 & 0 & -2 - \lambda \end{matrix} \right| = 0 \\ \lambda^3 - 2 \lambda^2 - 15 \lambda + 36 = 0 \\ (\lambda - 3)^2 (\lambda + 4) = 0$ .

This implies λ = 3 or λ = -4.

To determine the eigenspace of A corresponding to λ = 3, substitute λ = 3 into the system of equations $(A - \lambda I) \vec{x} = \vec{0}$ . Assuming $\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}$ , it follows that:

$\begin{bmatrix}2 & 6 & 2 \\ 0 & -4 & -8 \\ 1 & 0 & -5 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ …………………………………………………………………………………………………………………………………………………… (2)

By applying some elementary row operations, it can be easily shown that the solution set of (2) is equal to the solution set of (3) below.

$\begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ ………………………………………………………………………………………………………………………………………………………. (3)

Note that (3) corresponds to the system of linear equations $\left\{ \begin{matrix} x_1 & & -5 x_3 = 0 \\ & x_2 & + 2x_3 = 0 \end{matrix}$ .

The solution set of the system is $E_1 = \begin{Bmatrix} t \begin{bmatrix} 5 \\ -2 \\ 1 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ . It is an eigenspace of A corresponding to λ = 3.

If λ = -4, then $A - \lambda I = \begin{bmatrix}9 & 6 & 2 \\ 0 & 3 & -8 \\ 1 & 0 & 2 \end{bmatrix}$ , which is equivalent to $\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -8/3 \\ 0 & 0 & 0 \end{bmatrix}$ . If $\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}$ , the matrix corresponds to the system of linear equations represented as $\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -8/3 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ . The system has x₃ as the only free variable. So, if we set x₃ = 3t for any $t \in \mathbb{R}$ then x₂ = 8t and x₁ = -6t. As a consequence, the eigenspace of A corresponding to λ = -4 is $E_2 = \begin{Bmatrix} t \begin{bmatrix} -6 \\ 8 \\ 3 \end{bmatrix} \Bigr| t \in \mathbb{R} \end{Bmatrix}$ .

On 24/04/2022

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