Lesson 30 - Matrix Methods for Linear Homogeneous Systems

Definition: Let $A = [a_{i j}]_{n \times n}$ be a constant matrix. The eigenvalues of $A$ are values of $λ$ that satisfy $(A - λ I) \vec{v} = \vec{0}$ . The corresponding nontrivial solutions, $\vec{v}$ are called eigenvectors of $A$ associated with $λ$ .

Theorem: Suppose, $A$ , then $n \times n$ constant matrix has $n$ linearly independent eigenvectors $\vec{v_{1}}, \vec{v_{2}}, \dots, \vec{v_{n}}$ . If $λ_{i}$ are the corresponding eigenvalues (for $i = 1, 2, \dots, n$ ), then the set
{ $\vec{v_{1}} e^{λ_{1} t}, \vec{v_{2}} e^{λ_{2} t}, \dots, \vec{v_{n} e^{λ_{n} t}}$ } forms a fundamental set of solutions of the equations $\vec{x} (t) = \vec{A} x$ (t).

The general solution is

$\vec{x (t)} = c_{1} \vec{v_{1} e^{λ_{1} t}} + c_{2 \vec{v_{2}}} e^{λ_{2} t} + \dots + c_{n} \vec{v_{n}} e^{λ_{n} t}$

What does this all mean?

An eigenvalue is a special value that helps us understand how a matrix behaves when it acts on a vector. An eigenvector is a collection of those eigenvalues. These values are specific to the matrix that we are working with. They allow us to stretch or shrink the matrix without affecting the direction of the vector components.

For any matrix, we can find these special vectors and their corresponding values.

For differential equations, we deal with equations that change over time. These eigenvalues and eigenvectors help us solve these equations more easily. Let's take a look at an example.

Example

For the DE: $\vec{x} (t) = \vec{A} x (t)$ where $A = | \begin{matrix} 2 & - 3 \\ 1 & - 2 \end{matrix} |$ .

a) Find the eigenvalues $λ$ and the corresponding eigenvectors.
b) Write the general solution.

In order to find the eigenvalues, we take the determinant of $A - λ I$ , where $I$ is the identity matrix $| \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} |$ . We find that $λ I$ is a scalar-matrix multiplication, where we multiply each component of the matrix $I$ by the value $λ$ . This gives us $λ I = | \begin{matrix} λ & 0 \\ 0 & λ \end{matrix} |$

First, let's calculate $A - λ I$ .

A - λ I = | \begin{matrix} 2 & - 3 \\ 1 & - 2 \end{matrix} | - | \begin{matrix} λ & 0 \\ 0 & λ \end{matrix} |

= | \begin{matrix} 2 - λ & - 3 \\ 1 & - 2 - λ \end{matrix} |

Now, we find the determinant of the matrix. This is essentially the cross product of the matrix. So,

det | \begin{matrix} 2 - λ & - 3 \\ 1 & - 2 - λ \end{matrix} | = 0

(2 - λ) (- 2 - λ) - (- 3) (1) = 0

- 4 + λ^{2} + 3 = 0

λ^{2} + 1 = 0

λ = \pm 1

These are the eigenvalues of the matrix $A$ .

To find the eigenvectors, we have $(A - λ I) \vec{v} = \vec{0}$ , let $\vec{v} = | \begin{matrix} v_{1} \\ v_{2} \end{matrix} |$ .
For $λ = 1$ , we have

| \begin{matrix} 1 & - 3 \\ 1 & - 3 \end{matrix} | | \begin{matrix} v_{1} \\ v_{2} \end{matrix} | = 0

This comes from plugging in $λ = 1$ into the matrix that we found above for $A - λ I$ . Now, we have a linear system of equation in the form of vector-matrix multiplication. Let's convert this to notation that will allow us to solve for $v_{1}$ and $v_{2}$ .

v_{1} - 3 v_{2} = 0

v_{1} - 3 v_{2} = 0

If we simply let $v_{2} = s$ , we can solve for $v_{1}$ . $v_{1} = 3 s$ .

So now, we have an eigenvector

\vec{v} = | \begin{matrix} 3 s \\ s \end{matrix} | = s | \begin{matrix} 3 \\ 1 \end{matrix} |

Let's do the same for $λ = - 1$ .

| \begin{matrix} 3 & - 3 \\ 1 & - 1 \end{matrix} | | \begin{matrix} v_{1} \\ v_{2} \end{matrix} | = 0

3 v_{1} - 3 v_{2} = 0

v_{1} - v_{2} = 0

Again, $v_{1} = s$ . Therefore, $v_{2} = s$ . So our eigenvector is

\vec{v} = | \begin{matrix} s \\ s \end{matrix} | = s | \begin{matrix} 1 \\ 1 \end{matrix} |

So now, our solution is

\vec{x} (t) = c_{1} | \begin{matrix} 3 \\ 1 \end{matrix} | e^{t} + c_{2} | \begin{matrix} 1 \\ 1 \end{matrix} | e^{- t}

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