# Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization

## Problem 667

(a) Find all solutions of the linear dynamical system
$\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x},$ where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a function of the variable $t$.

(b) Solve the linear dynamical system
$\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}$ with the initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$.

## Solution.

### (a) Find all solutions of the linear dynamical system $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x}$

Note that the given system is
$\begin{bmatrix} \frac{\mathrm{d}{x_1}}{\mathrm{d} t} \\[6pt] \frac{\mathrm{d}{x_2}} {\mathrm{d} t} \end{bmatrix} =\begin{bmatrix} x_1 \\ 3x_2 \end{bmatrix}.$ Thus, we have two uncoupled differential equations
\begin{align*}
\frac{\mathrm{d}{x_1}}{\mathrm{d} t} &=x_1 \6pt] \frac{\mathrm{d}{x_2}}{\mathrm{d} t} &=3x_2. \end{align*} The solutions to these differential equations are \begin{align*} x_1(t)&=e^t x_1(0)\\ x_2(t)&=e ^{3t} x_2(0). \end{align*} Thus we see that \[\mathbf{x}(t)=\begin{bmatrix} e^t x_1(0) \\[6pt] e ^{3t} x_2(0) \end{bmatrix}.

### (b) Solve the linear dynamical system $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}$

Let $A=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}$.
Then the matrix $A$ has eigenvalues $1, 3$ and corresponding eigenvectors are
$\begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 \\ 1 \end{bmatrix},$ respectively.
(See the post Diagonalize a 2 by 2 Symmetric Matrix for details.)

Thus, if we put $S=\begin{bmatrix} 1 & -1\\ 1& 1 \end{bmatrix}$, then $A$ is diagonalizable by $S$. That is,
$S^{-1}AS=D,$ where
$D=\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}.$

Substituting $A=SDS^{-1}$ into the given system, we have
\begin{align*}
\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}&=(SDS^{-1})\mathbf{x}\6pt] \Leftrightarrow \quad S^{-1}\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}&=(DS^{-1})\mathbf{x}\\[6pt] \Leftrightarrow \quad \frac{\mathrm{d}(S^{-1} \mathbf{x})}{\mathrm{d}t}&=(DS^{-1})\mathbf{x}. \end{align*} Let \mathbf{u}(t)=S^{-1}\mathbf{x}. Then we obtain the system \[\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{u}. We solved this system in part (a) and the general solution is given by
$\mathbf{u}(t)=\begin{bmatrix} e^t u_1(0) \\[6pt] e^{3t} u_2(0) \end{bmatrix}.$

We determine the values of $u_1(0)$ and $u_2(0)$ using the given initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$.
We have
$\begin{bmatrix} u_1(0) \\ u_2(0) \end{bmatrix}=\mathbf{u}(0)=S^{-1} \mathbf{x}(0)=\frac{1}{2}\begin{bmatrix} 1 & 1\\ -1& 1 \end{bmatrix}\begin{bmatrix} 1 \\ 3 \end{bmatrix}=\begin{bmatrix} 2 \\ 1 \end{bmatrix}.$ (Note that $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ is the coordinate vector of $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ with respect to the eigenbasis $\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.)

It follows that the solution of the original system is
\begin{align*}
\mathbf{x}=S\mathbf{u}=\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix}
\begin{bmatrix}
2e^t \6pt] e^{3t} \end{bmatrix} =2e^t\begin{bmatrix} 1 \\ 1 \end{bmatrix}+ e^{3t}\begin{bmatrix} -1 \\ 1 \end{bmatrix}. \end{align*} ### Another Solution of (b) In this solution, we use the following theorem. Theorem. Let A be a diagonalizable n\times n matrix. Let \{\mathbf{v}_1,\dots, \mathbf{v}_n\} be an eigenbasis for A, with associated eigenvalues \lambda_1, \dots, \lambda_n. Then the general solution of the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x} is
$\mathbf{x}(t)=c_1 e^{\lambda_1 t}\mathbf{v}_1+\cdots +c_n e^{\lambda_n t}\mathbf{v}_n,$ where $c_1, \dots, c_n$ are arbitrary complex numbers.

As in the above solution, we know that the matrix $A=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}$ has eigenvalues $1, 3$ and corresponding eigenvectors are
$\begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 \\ 1 \end{bmatrix},$ respectively.

So the formula in the theorem yields the general solution
$\mathbf{x}(t)=c_1 e^{t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}+c_2 e^{3t}\begin{bmatrix} -1 \\ 1 \end{bmatrix},$ where $c_1, c_2$ are constants.

Since the initial is $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$, we have
\begin{align*}
\begin{bmatrix}
1 \\
3
\end{bmatrix}=c_1\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
-1 \\
1
\end{bmatrix}.
\end{align*}
Solving this system, we obtain $c_1=2$ and $c_2=1$.
Thus, the solution of the linear dynamical system with the given initial value is
$\mathbf{x}(t)=2 e^{t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}+e^{3t}\begin{bmatrix} -1 \\ 1 \end{bmatrix}.$

##### Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less. (a) Prove that the set \$\{ 1...