# Homogeneous Linear Systems

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 5.4 – Differential Operators and the Elimination Method for Systems
• Section 9.4 -- Linear Systems in Nominal Form
• Section 9.5 -- Homogeneous Linear Systems with Constant Coefficients
• Section 9.6 -- Complex Eigenvalues

## Expected Educational Results

• Objective 16–1: I can homogeneous linear systems using eigenvalues and eigenvectors.

## Homogeneous Linear Systems

### System of $n$ Linear DEs

Given the following homogeneous system of $n$ linear DEs:

$\pmb{x}^{\,\prime}(t)=\pmb{A}\pmb{x}(t)$

$\displaystyle \Rightarrow \left[ \begin{array}{c} x_1^{\,\prime}(t) \\ x_2^{\,\prime}(t) \\ \vdots \\ x_n^{\,\prime}(t) \end{array} \right]=\pmb{A}\left[ \begin{array}{c} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{array} \right]$

where $\displaystyle \pmb{A}=\left[a_{ij}\right]$ is the $n\times n$ coefficient matrix of [constant] coefficients of $x_{ij}$.

Let’s try the functions $\vec{x}(t)=\vec{u}e^{\lambda t}$, where $\vec{u}$ is a vector of arbitrary constants and $\lambda$ is a number (it is called the eigenvalue of the coefficient matrix), as solutions to the homogeneous system of linear DEs (see CPT_10c_Characteristic_Equation.html). So,

$\lambda\vec{u}e^{\lambda t}=\pmb{A}\vec{u}e^{\lambda t}$

This simplifies to:

$\lambda\vec{u}=\pmb{A}\vec{u}$

We need to solve for $\lambda$, i.e., root of the characteristic equation. So,

$\pmb{A}\vec{u}-\lambda\vec{u}=\vec{0}$

$\Rightarrow\left(\pmb{A}-\lambda\pmb{I}\right)\vec{u}=\vec{0}$, where $\vec{u}\ne\vec{0}$

### System of Two Linear DEs

#### Solve for eigenvalues, $\lambda$:

$\left(\pmb{A}-\lambda\pmb{I}\right)\vec{u}=0$

$\displaystyle \Rightarrow \left(\left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]-\lambda\left[ \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array} \right]\right)\left[ \begin{array}{c} u_1 \\ u_2 \end{array} \right]=\vec{0}$, $\vec{u}\ne \vec{0}$

The above has a solution only if

$\displaystyle \text{det}\left(\left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]-\lambda\left[ \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array} \right]\right)=0$

$\displaystyle \Rightarrow \text{det}\left(\left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]-\left[ \begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array} \right]\right)=$

$\displaystyle \Rightarrow \text{det}\left(\left[ \begin{array}{ll} a_{11}-\lambda & a_{12} \\ a_{21} & a_{22}-\lambda \end{array} \right]\right)=0$

$\displaystyle \Rightarrow (a_{11}-\lambda)(a_{22}-\lambda) - a_{12}a_{21} =0$

The $\displaystyle 2^{\text{nd}}$-degree polynomial in $\lambda$ below is called the characteristic polynomial.

$\displaystyle \Rightarrow \lambda^2 - (a_{11}+a_{22})\lambda - a_{12}a_{21} = 0$

Solve for $\lambda_1$ and $\lambda_2$ using any valid algebraic method.

#### Solve for eigenvectors:

Use $\lambda_1$ to solve for eigenvector $\vec{u}_1=\left[ \begin{array}{l} u_{11} \\ u_{12} \end{array} \right]$:

$\left(\pmb{A}-\lambda_1\pmb{I}\right)\vec{u}_1=0$, $\vec{u}_1\ne \vec{0}$

$\displaystyle \Rightarrow \left(\left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]-\lambda_1\left[ \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array} \right]\right)\vec{u}_1=\left[ \begin{array}{ll} a_{11}-\lambda_1 & a_{12} \\ a_{21} & a_{22}-\lambda_1 \end{array} \right]\vec{u}_1=\vec{0}$

Using matrix multiplication [there are other procedures to solve for the components $\vec{u}_1$]:

$\Rightarrow\left[ \begin{array}{l} (a_{11}-\lambda_1)u_{11} + a_{12}u_{12} \\ a_{21}u_{11} + (a_{22}-\lambda_1)u_{12} \end{array} \right]=\left[ \begin{array}{l} 0 \\ 0 \end{array} \right]$

Solve the above system of two equations and two unknowns for $u_{11}$ and $u_{12}$.

So, the eigenvector that corresponds to the eigenvalue, $\lambda_1$ is any vector for which

$\vec{u}_1=\left[ \begin{array}{l} u_{11} \\ u_{12} \end{array} \right]$ or any multiple of this vector.

Similarly, use $\lambda_2$ to solve for eigenvector $\vec{u}_2$:

$\left(\pmb{A}-\lambda_2\pmb{I}\right)\vec{u}_2=0$, $\vec{u}_2\ne \vec{0}$

Verify $\lambda_1$ and $\vec{u}_1$ by showing the equivalence:

$\pmb{A}\vec{u}_1=\lambda_1\vec{u}_1$

$\Rightarrow\left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]\left[ \begin{array}{l} u_{11} \\ u_{12} \end{array} \right]=\lambda_1\left[ \begin{array}{l} u_{11} \\ u_{12} \end{array} \right]$

Similarly, verify $\lambda_2$ and $\vec{u}_2$.