# L07
###### tags: `微分方程特論`
###### date: `15 Apr 2020`
（第3次線上上課）
# Review of Lesson 6
Sec. 5.4: The Eigenvalue Method for Homogeneous Systems<br>Sec. 5.6: Multiple Eigenvalue Solution</br>
---
:film_frames: [202004015_1.MTS](https://reurl.cc/20zMp4) 2:46
1. We have been solving linear system of form $$\dfrac{d\mathbf{x}}{dt}=\mathsf{A}\mathbf{x},\tag{1}$$where $\mathsf{A}$ is a matrix with constant coefficients.
2. We found that $\mathbf{x}(t)=\mathbf{v}e^{\lambda t}$ is a solution of System (1), where $\lambda$ is an eigenvalue of $\mathsf{A}$ , and $\mathbf{v}$ is the eigenvector associated with $\lambda$.
3. We discussed three cases about $\lambda$:
1. [Case 1: distinct real eigenvalues]()
2. [Case 2: complex eigenvalues](https://hackmd.io/@ulynx/HykQWysPI#Case-2-complex-eigenvalues)
3. Case 3: multiple eigenvalues
- definition
::: success
- $k$ is called the **multiplicity** of eigenvalue $\lambda$;
- $p$ is the number of linearly independent solutions associated with $\lambda$;
- $d=k-p$ is called the **defect** of eigenvalue $\lambda$.
:::
- [Case 3a: complete eigenvalue](https://hackmd.io/1h05FPs9RlmxaLp7xoHl1w?view#Case-3a-%E2%80%9Ccomplete%E2%80%9D-multiple-real-eigenvalues) $d=0$
- [Case 3b: defective eigenvalue](https://hackmd.io/1h05FPs9RlmxaLp7xoHl1w?view#Case-3b-%E2%80%9Cdefective%E2%80%9D-multiple-real-eigenvalues) $d>0$
- how to find generalized eigenvectors?
# Lesson 07
Sec. 5.6: Multiple Eigenvalue Solution
---
:film_frames: [202004015_1.MTS](https://reurl.cc/20zMp4) 13:06
For the eigenvalue $\lambda$ of $k=3$ and $d=1$ ($p=2$), we can find two linearly independent eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$, then $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}$ and $\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}$ are linearly independent. However, if we try $$(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\mathbf{v}_1\quad\text{or}\quad(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\mathbf{v}_2$$ just as we did before, we will fail to find the third L.I. eigenvector.
For the sake of greater generality, we try $$(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\alpha \mathbf{v}_1+\beta\mathbf{v}_2\tag{2}$$ with constants $\alpha$, $\beta$ underdetermined, and $$\mathbf{x}_3=(\alpha \mathbf{v}_1t+\beta\mathbf{v}_2t+\mathbf{v}_3)e^{\lambda t}.$$
### Example of Case 3b with $k=3$, $d=1$
- :film_frames: [202004015_1.MTS](https://reurl.cc/20zMp4) 17:08
::: warning
Find a general solution of $$\mathbf{x}'=\underbrace{\begin{pmatrix}4&3&1\\-4&-4&-2\\8&12&6\end{pmatrix}}_\mathsf{A}\mathbf{x}.\tag{3}$$
:::
#### Solution
From $$\begin{align}0&=\begin{vmatrix}4-\lambda&3&1\\-4&-4-\lambda&-2\\8&12&6-\lambda\end{vmatrix}\\&=(4-\lambda)(-4-\lambda)(6-\lambda)-48-48+8(4+\lambda)+24(4-\lambda)+12(6-\lambda)\\&=\lambda^3-6\lambda^2+12\lambda-8\\&=(\lambda-2)^3\end{align}$$ we find the eigenvalue $\lambda=2$ of multiplicity $k=3$.
For $\lambda=2$, the eigenvector equation $$\begin{pmatrix}2&3&1\\-4&-6&-2\\8&12&4\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\tag{4}$$ only gives one constraint $2a+3b+c=0$ on $a,b,c$, leaving two of them arbitrary.
- If we choose $a=1$ and $b=0$, then $c=-2$ and thus one eigenvector associated with $\lambda=2$ is ==$\mathbf{v}_1=\begin{pmatrix}1&0&-2\end{pmatrix}^T$==,
- If we choose $a=0$ and $b=1$, then $c=-3$ and thus another eigenvector associated with $\lambda=2$ is ==$\mathbf{v}_1=\begin{pmatrix}0&1&-3\end{pmatrix}^T$==.
We found two L.I. solutions of System (3), so the defect of $\lambda=2$ is $d=1$.
Now we consider $\mathbf{v}_3=\begin{pmatrix}a&b&c\end{pmatrix}^T$ and assume it satisfies the equation $(\mathsf{A}-2\mathsf{I})\mathbf{v}_3=\alpha \mathbf{v}_1+\beta\mathbf{v}_2$: $$\begin{pmatrix}2&3&1\\-4&-6&-2\\8&12&4\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\alpha\begin{pmatrix}1\\0\\-2\end{pmatrix}+\beta\begin{pmatrix}0\\1\\-3\end{pmatrix}$$ or $$\left\{\begin{array}{rcrcrcl}2a&+&3b&+&c&=&\alpha\\-4a&-&6b&-&2c&=&\beta\\8a&+&12b&+&4c&=&-2\alpha-3\beta\end{array}\right..\tag{5}$$ From the first and the second rows we know that $\beta=-2\alpha$; also from the second and the third rows we know $-2\alpha=\beta$. If we chhose $\alpha=1$, then $\beta=-2$. Equations (5) becomes $$\left\{\begin{array}{rcrcrcl}2a&+&3b&+&c&=&1\\-4a&-&6b&-&2c&=&-2\\8a&+&12b&+&4c&=&4\end{array}\right.,\tag{5}$$ and yields $2a+3b+c=1$. This time we choose $a=b=0$, so that $c=1$ and $\mathbf{v}_3=\begin{pmatrix}0&0&1\end{pmatrix}^T$.
The third L.I. solution is $$\begin{align}\mathbf{x}_3(t)&=(\alpha \mathbf{v}_1t+\beta\mathbf{v}_2t+\mathbf{v}_3)e^{2t}\\&=te^{2t}\begin{pmatrix}1\\0\\-2\end{pmatrix}-2te^{2t}\begin{pmatrix}0\\1\\-3\end{pmatrix}+e^{2t}\begin{pmatrix}0\\0\\1\end{pmatrix}\\&=e^{2t}\begin{pmatrix}t\\-2t\\4t+1\end{pmatrix}.\end{align}$$
Therefore, the general solution of System (3) is $$\begin{align}\mathbf{x}(t)&=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+c_3\mathbf{x}_3(t)\\&=c_1e^{2t}\begin{pmatrix}1\\0\\-2\end{pmatrix}+c_2e^{2t}\begin{pmatrix}0\\1\\-3\end{pmatrix}+c_3e^{2t}\begin{pmatrix}t\\-2t\\4t+1\end{pmatrix}\\&=e^{2t}\begin{pmatrix}c_1+c_3t\\c_2-2c_3t\\-2c_1-3c_2+4c_3t+c_3\end{pmatrix}\end{align}\tag{6}$$ where $c_1,c_2,c_3$ are arbitrary real constants. $\blacksquare$
#### MATLAB
**Input**
``` v
A=[4 3 1;-4 -4 -2;8 12 6];
[V D]=eig(A)
```
**Output**
```
[V D]=eig(A)
V =
0.21822 -0.21822 -0.24339
-0.43644 0.43644 0.44886
0.87287 -0.87287 -0.85982
D =
Diagonal Matrix
2.0000 0 0
0 2.0000 0
0 0 2.0000
```
### Summary of generalized eigenvector and L.I. solutions
| Multiplicity | Defect |Eigenvec. eqn.| L.I. solutions |
|:------------:|:-----------------:|:------:| -------- |
| $k=2$ | $d=1$ | $\require{color}(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_2=\mathbf{0}$ | $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\color{red}{\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2)e^{\lambda t}}$ |
| $k=2$ | $d=0$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}$ | $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}$ |
| Multiplicity | Defect |Eigenvec. eqn.| L.I. solutions |
|:------------:|:------:|:------:| ------------------------------------------------------------------------------------------------------ |
| $k=3$ | $d=2$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^3\mathbf{v}_3=\mathbf{0}$ | $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2)e^{\lambda t}\\\color{red}{\mathbf{x}_3(t)=(\frac{1}{2}\mathbf{v}_1t^2+\mathbf{v}_2t+\mathbf{v}_3)e^{\lambda t}}$ |
| $k=3$ | $d=1$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_3=\mathbf{0}$|$\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}\\\color{c}{\mathbf{x}_3(t)=\left(\alpha \mathbf{v}_1t+\beta\mathbf{v}_2t+\mathbf{v}_3\right)e^{\lambda t}}$ |
| $k=3$ | $d=0$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\mathbf{0}$ |$\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}\\\mathbf{x}_3(t)=\mathbf{v}_3e^{\lambda t}$ |
:film_frames: [202004015_2.MTS](https://reurl.cc/GV4exv) 00:00
| Multiplicity | Defect |Eigenvec. eqn.| L.I. solutions |
|:------------:|:------:|:------:| ------------------------------------------------------------------------------------------------------ |
| $k=4$ | $d=3$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^3\mathbf{v}_3=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^4\mathbf{v}_4=\mathbf{0}$ | $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2)e^{\lambda t}\\\mathbf{x}_3(t)=(\frac{1}{2}\mathbf{v}_1t^2+\mathbf{v}_2t+\mathbf{v}_3)e^{\lambda t}\\\color{red}{\mathbf{x}_4(t)=(\frac{1}{3!}\mathbf{v}_1t^3+\frac{1}{2!}\mathbf{v}_2t^2+\mathbf{v}_3t+\mathbf{v}_4)e^{\lambda t}}$ |
| $k=4$ | $d=2$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_3=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_4=\mathbf{0}$| $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}\\\color{coral}{\mathbf{x}_3(t)=\left(\alpha \mathbf{v}_1t+\beta\mathbf{v}_2t+\mathbf{v}_3\right)e^{\lambda t}\\\mathbf{x}_4(t)=\left(\gamma \mathbf{v}_1t+\delta\mathbf{v}_2t+\mathbf{v}_4\right)e^{\lambda t}}$ |
| $k=4$ | $d=1$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_4=\mathbf{0}$ |$\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}\\\mathbf{x}_3(t)=\mathbf{v}_3e^{\lambda t}\\\color{coral}{\mathbf{x}_4(t)=(\alpha\mathbf{v}_1t+\beta\mathbf{v}_2t+\gamma\mathbf{v}_3t+\mathbf{v}_4)e^{\lambda t}}$ |
| $k=4$ | $d=0$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_4=\mathbf{0}$ |$\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}\\\mathbf{x}_3(t)=\mathbf{v}_3e^{\lambda t}\\\mathbf{x}_4(t)=\mathbf{v}_4e^{\lambda t}$ |
### Example of Case2b $k=3$,
- :film_frames: [202004015_2.MTS](https://reurl.cc/GV4exv) 08:00
- p.403, Example 6 of Sec. 5.6
::: warning
Find a general solution of the system $$\mathbf{x}'=\begin{pmatrix}0&0&1&0\\0&0&0&1\\-2&2&-3&1\\2&-2&1&-3\end{pmatrix}\mathbf{x}.\tag{7}$$
:::
#### Solution
The characteristic equation of the coefficient matrix $\mathsf{A}$ in System (7) is $$\lambda^4+6\lambda^3+12\lambda^2+8\lambda=\lambda(\lambda+2)^3=0.$$ Thus $\mathsf{A}$ has the distinct eigenvalue $\lambda_1=0$ and the triple eigenvalue $\lambda_2=2$.
For $\lambda_1=0$, the eigenvalue equation is $$\begin{pmatrix}0&0&1&0\\0&0&0&1\\-2&2&-3&1\\2&-2&1&-3\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}.$$ The fisrt two rows give $c=d=0$, then the last two rows tield $a=b$. Thus ==$\mathbf{v}_1=\begin{pmatrix}1&1&0&0\end{pmatrix}^T$== is an eigenvector associated with $\lambda_1=0$.
For $\lambda_2=-2$, the eigenvalue equation is $$\begin{pmatrix}2&0&1&0\\0&2&0&1\\-2&2&-1&1\\2&-2&1&-1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}.$$ The last two equations are redundant because they are the differences of first two ones, while the first ones give $2a+c=0$ and $2b+d=0$.
- The choice $a=1$, $b=0$ yields $c=-2$, $d=0$ and thereby the eigenvector ==$\mathbf{v}_2=\begin{pmatrix}1&0&-2&0\end{pmatrix}^T$==.
- The choice $a=0$, $b=1$ yields $c=0$, $d=-2$ and thereby the eigenvector ==$\mathbf{v}_3=\begin{pmatrix}0&1&0&-2\end{pmatrix}^T$==.
Both of $\mathbf{v}_2$ and $\mathbf{v}_3$ are linearly independent eigenvectors of $\mathsf{A}$ associated with the eigenvalue $\lambda_2=-2$. Because $\lambda_2=-2$ has a defect $1$, we we need a generalized eigenvector of rank $2$.
From $$\begin{pmatrix}2&0&1&0\\0&2&0&1\\-2&2&-1&1\\2&-2&1&-1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\alpha\begin{pmatrix}1\\0\\-2\\0\end{pmatrix}+\beta\begin{pmatrix}0\\1\\0\\-2\end{pmatrix},$$ we obtain the system $$\left\{\begin{array}{rcl}2a+c&=&\alpha\\2b+d&=&\beta\\-2a+2b-c+d&=&-2\alpha\\2a-2b+c-d&=&-2\beta\end{array}\right.$$
Comparing the third equation and the first equation substracted by the second one suggests that $-2\alpha=-(\alpha-\beta)$ or $\alpha=-\beta$. We choose $\alpha=1$ and then $\beta=-1$. The equation becomes $$\begin{pmatrix}2&0&1&0\\0&2&0&1\\-2&2&-1&1\\2&-2&1&-1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}1\\-1\\-2\\2\end{pmatrix}.$$
The last two equations are still redundant because they are the differences of first two ones, while the first ones give $2a+c=1$ and $2b+d=-1$. We choose $a=0$, $b=0$, so that $c=1$, $d=-1$ and thus ==$\mathbf{v}_4=\begin{pmatrix}0&0&1&-1\end{pmatrix}^T$== is still another eigenvector associated with eigenvalue $\lambda_2$.
Therefore the linear independent solutions are $$\begin{align}\mathbf{x}_1(t)&=\mathbf{v}_1e^{0t}=\begin{pmatrix}1&1&0&0\end{pmatrix}^T\\\mathbf{x}_2(t)&=\mathbf{v}_2e^{-2t}=\begin{pmatrix}1&0&-2&0\end{pmatrix}^Te^{-2t}\\\mathbf{x}_3(t)&=\mathbf{v}_3e^{-2t}=\begin{pmatrix}0&1&0&-2\end{pmatrix}^Te^{-2t}\\\mathbf{x}_4(t)&=\left(\alpha\mathbf{v}_2t+\beta\mathbf{v}_3t+\mathbf{v}_4\right)e^{-2t}\\&=\begin{pmatrix}t&-t&-2t+1&2t-1\end{pmatrix}^Te^{-2t}\end{align}$$ and the general solution of System (7) is $$\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+c_3\mathbf{x}_3(t)+c_4\mathbf{x}_4(t),$$ where where $c_1,c_2,c_3$ are arbitrary real constants. $\blacksquare$
Sec. 5.7: Matrix Exponentials and Linear Systems
---
### 1) Matrix Differential Equation
:film_frames: [202004015_3.MTS](https://reurl.cc/ZO9Wmp) 00:00
Consider a $n\times n$ homogeneous linear system $$\mathbf{x}'=\mathsf{A}\mathbf{x},\tag{8}$$ where $\mathbf{x}$ is a vector and $\mathsf{A}$ is an $n\times n$ matrix (that may depend on $t$). If System (8) has $n$ solutions $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_n$, then the matrix $$\mathsf{X}(t)=\begin{pmatrix}|&|&|&|\\\mathbf{x}_1&\mathbf{x}_2&\cdots&\mathbf{x}_n\\|&|&|&|\end{pmatrix}_{n\times n}\tag{9}$$ satisifes the **matrix differential equation** $$\mathsf{X}'=\mathsf{A}\mathsf{X}.\tag{10}$$ Equation (10) is also a linear system of equations represented by matrix form.
### 2) Fundamental Matrix Solutions
If $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_n$ are $n$ linearly independent solutions of System (8), then $\{\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_n\}$ is called a **fundamental set of solutions** and the matrix $$\mathsf{\Phi}(t)=\begin{pmatrix}|&|&|&|\\\mathbf{x}_1&\mathbf{x}_2&\cdots&\mathbf{x}_n\\|&|&|&|\end{pmatrix}_{n\times n}\tag{11}$$ is called a **fundamental matrix** of System (8). With this definition, a general solution of System (8) can be written in $$\begin{align}\mathbf{x}(t)&=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+\cdots+c_n\mathbf{x}_n(t)\\&=\begin{pmatrix}|&|&|&|\\\mathbf{x}_1(t)&\mathbf{x}_2(t)&\cdots&\mathbf{x}_n(t)\\|&|&|&|\end{pmatrix}\begin{pmatrix}c_1\\c_2\\\vdots\\c_n\end{pmatrix}=\mathsf{\Phi}(t)\mathbf{c},\end{align}\tag{12}$$ where $\mathbf{c}=\begin{pmatrix}c_1&c_2&\cdots&c_n\end{pmatrix}^T$ is an arbitrary *constant* vector.
Given an initial condition, System (8) forms an initial value problem (I.V.P.) $$\mathbf{x}'=\mathsf{A}\mathbf{x},\quad \mathbf{x}(0)=\mathbf{x}_0.\tag{13}$$ Then, as we just mentioned, a general solution is $\mathbf{x}(t)=\mathsf{\Phi}(t)\mathbf{c}$, so we have $\mathbf{x}(0)=\mathsf{\Phi}(0)\mathbf{c}=\mathbf{x}_0$ and thus $$\mathbf{c}=\mathsf{\Phi}(0)^{-1}\mathbf{x}_0\tag{14}$$
When we substitute (14) in (12), we see that the solution of the I.V.P. (13) is given by $$\mathbf{x}(t)=\mathsf{\Phi}(t)\overbrace{\mathsf{\Phi}(0)^{-1}\mathbf{x}_0}^\mathbf{c}.\tag{15}$$
It is always valid to write (14) and (15) since the $n$ columns of $\mathsf{\Phi}$ are linearly independent, which implies that $\mathsf{\Phi}$ is invertible, i.e., $\mathsf{\Phi}^{-1}$ exists. Actually, the determinant of $\mathsf{\Phi}$ is the Wronskian of $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_n$: $$\det\mathsf{\Phi}=\begin{vmatrix}|&|&|&|\\\mathbf{x}_1&\mathbf{x}_2&\cdots&\mathbf{x}_n\\|&|&|&|\end{vmatrix}=W(\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_n;t).$$ The Wronskian of linearly independent vectors never vanishes and $\det\mathsf{\Phi}\neq 0$ guarantees the exisitence of $\mathsf{\Phi}^{-1}$.
**Note**: In equations (13) to (15), we may shift the value from which initial conditions are taken from $0$ and any real number $t_0$.
### Example 1
- :film_frames: [202004015_3.MTS](https://reurl.cc/ZO9Wmp) 12:19
- p.409, Example 1 of Sec. 5.7
::: warning
Find a fundamental matrix for the system $$\left\{\begin{array}{l}x'=4x+2y\\y'=3x-y\end{array}\right.,\tag{16}$$ then use it to find the solution of (13) that satisfies the initial conditions $x(0) = 1$, $y(0) = −1$.
:::
#### Solution
Let $$\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix},\quad\mathbf{x}_0=\begin{pmatrix}1\\-1\end{pmatrix},\quad \text{and }\mathsf{A}=\begin{pmatrix}4&2\\3&-1\end{pmatrix}.$$ Then System (16) can be written in $\mathbf{x}'=\mathsf{A}\mathbf{x}$.
First we solve the secular equation of $\mathsf{A}$: $$\begin{vmatrix}4-\lambda&2\\3&-1-\lambda\end{vmatrix}=\lambda^2-3\lambda-10=(\lambda-5)(\lambda+2)=0.$$ Thus $\lambda_1=5$ and $\lambda_2=-2$ are eigenvalues of $\mathsf{A}$
For $\lambda_1=5$, the equation $(\mathsf{A}-5\mathsf{I})\mathbf{v}_1=\mathbf{0}$ or $$\begin{pmatrix}-1&2\\3&-6\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ suggests the choice $a=2$, $b=1$ and thereby the eigenvector $\mathbf{v}_1=\begin{pmatrix}2&1\end{pmatrix}^T$
For $\lambda_2=-2$, the equation $(\mathsf{A}+2\mathsf{I})\mathbf{v}_2=\mathbf{0}$ or $$\begin{pmatrix}6&2\\3&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ suggests the choice $a=1$, $b=-3$ and thereby the eigenvector $\mathbf{v}_2=\begin{pmatrix}1&-3\end{pmatrix}^T$.
Therefore, the two L.I. solutions are $\mathbf{x}_1=\begin{pmatrix}2e^{5t}\\e^{5t}\end{pmatrix}$ and $\mathbf{x}_2=\begin{pmatrix}e^{-2t}\\-3e^{-2t}\end{pmatrix}$ and the fundamental matrix $$\boxed{\mathsf{\Phi}=\begin{pmatrix}2e^{5t}&e^{-2t}\\e^{5t}&-3e^{-2t}\end{pmatrix}}.$$
The formula in (15) gives $$\begin{align}\mathbf{x}(t)&=\begin{pmatrix}2e^{5t}&e^{-2t}\\e^{5t}&-3e^{-2t}\end{pmatrix}\begin{pmatrix}2&1\\1&-3\end{pmatrix}^{-1}\begin{pmatrix}1\\-1\end{pmatrix}\\&=-\dfrac{1}{7}\begin{pmatrix}2e^{5t}&e^{-2t}\\e^{5t}&-3e^{-2t}\end{pmatrix}\begin{pmatrix}-3&-1\\-1&2\end{pmatrix}\begin{pmatrix}1\\-1\end{pmatrix}\\&=\begin{pmatrix}2e^{5t}&e^{-2t}\\e^{5t}&-3e^{-2t}\end{pmatrix}\begin{pmatrix}\frac{2}{7}\\\frac{3}{7}\end{pmatrix}\\&=\begin{pmatrix}\frac{4}{7}e^{5t}+\frac{3}{7}e^{-2t}\\\frac{2}{7}e^{5t}-\frac{9}{7}e^{-2t}\end{pmatrix}.\end{align}$$ Thus the solution of the original initial value problem is given by $$\boxed{x(t)=\tfrac{4}{7}e^{5t}+\tfrac{3}{7}e^{-2t},\quad y(t)=\tfrac{2}{7}e^{5t}-\tfrac{9}{7}e^{-2t}}.\tag{17}$$
### 3) Exponential Matrices
- :film_frames: [202004015_3.MTS](https://reurl.cc/ZO9Wmp) 21:09
We know that $x=e^{at}$ is a solution of $x'=ax$. It is natural to guess that $\mathsf{X}=e^{\mathsf{A}t}$ is a solution of $\mathsf{X}'=\mathsf{A}\mathsf{X}$. But what does $e^{\mathsf{A}t}$ mean?
Similar to the exponential series $$e^a=1+a+\dfrac{a^2}{2!}+\cdots+\dfrac{a^n}{n!}+\cdots,$$ the **exponential matrix** $e^{\mathsf{A}}$ is the $n\times n$ matrix defined by the series $$e^\mathsf{A}=\mathsf{I}+\mathsf{A}+\dfrac{\mathsf{A}^2}{2!}+\cdots+\dfrac{\mathsf{A}^n}{n!}+\cdots=\sum^{\infty}_{n=0}\dfrac{\mathsf{A}^n}{n!},\tag{18}$$ where $\mathsf{A}^0\equiv\mathsf{I}$ is the $n\times n$ identity matrix.
### Example 2
- :film_frames: [202004015_3.MTS](https://reurl.cc/ZO9Wmp) 24:29
- p.411, Example 2 of Sec. 5.7
::: warning
Let $\mathsf{A}=\begin{pmatrix}a&0\\0&b\end{pmatrix}.$ Calculate $e^{\mathsf{A}}$.
:::
#### Solution
We have $$\begin{align}\mathsf{A}^2&=\begin{pmatrix}a&0\\0&b\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}=\begin{pmatrix}a^2&0\\0&b^2\end{pmatrix},\\ \mathsf{A}^3&=\begin{pmatrix}a^2&0\\0&b^2\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}=\begin{pmatrix}a^3&0\\0&b^3\end{pmatrix},\\&\;\;\vdots\\ \mathsf{A}^n&=\begin{pmatrix}a^{n-1}&0\\0&b^{n-1}\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}=\begin{pmatrix}a^n&0\\0&b^n\end{pmatrix}.\end{align}$$
Therefore, $$\begin{align}e^{\mathsf{A}}&=\mathsf{I}+\mathsf{A}+\dfrac{\mathsf{A}^2}{2!}+\cdots+\dfrac{\mathsf{A}^n}{n!}+\cdots\\&=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}a&0\\0&b\end{pmatrix}+\dfrac{1}{2!}\begin{pmatrix}a^2&0\\0&b^2\end{pmatrix}+\cdots+\dfrac{1}{n!}\begin{pmatrix}a^n&0\\0&b^n\end{pmatrix}+\cdots\\&=\begin{pmatrix}1+a+\frac{a^2}{2!}+\cdots+\frac{a^n}{n!}+\cdots&0\\0&1+b+\frac{b^2}{2!}+\cdots+\frac{b^n}{n!}+\cdots\end{pmatrix}\\&=\boxed{\begin{pmatrix}e^a&0\\0&e^b\end{pmatrix}}.\end{align}$$ So the exponential of the diagonal $2\times2$ matrix $\mathsf{A}$ is obtained simply by exponentiating each diagonal element of $\mathsf{A}$. $\blacksquare$
We have an $n\times n$ analog of the previous result: if $\mathsf{D}=\text{diag}\left(d_1,d_2,\ldots,d_n\right)$ is an $n\times n$ diagonal matrix, then $e^\mathsf{D}=\text{diag}\left(e^{d_1},e^{d_2},\ldots,e^{d_n}\right)$.
### 4) Properties
:film_frames: [202004015_3.MTS](https://reurl.cc/ZO9Wmp) 28:39
1. $e^{\mathsf{O}_{n\times n}}=\mathsf{I}_{n},\tag{19}$
\
where $\mathsf{O}_{n\times n}$ is the $n\times n$ zero matrix and $\mathsf{I}_{n}$ is the $n\times n$ identity matrix.
:film_frames: [202004015_4.MTS](https://reurl.cc/d0LLvz) 00:00
2. $e^{\mathsf{A}+\mathsf{B}}=e^\mathsf{A}e^\mathsf{B}\tag{20}$
\
only if $\mathsf{A}$ and $\mathsf{B}$ commute, i.e. $\mathsf{A}\mathsf{B}=\mathsf{B}\mathsf{A}$. This is because $e^{\mathsf{A}+\mathsf{B}}=\mathsf{I}+(\mathsf{A}+\mathsf{B})+\dfrac{1}{2}(\mathsf{A}+\mathsf{B})^2+\cdots$ involves terms such as $\mathsf{A}^2+\mathsf{A}\mathsf{B}+\mathsf{B}\mathsf{A}+\mathsf{B}^2$, which does not equal to $\mathsf{A}^2+2\mathsf{A}\mathsf{B}+\mathsf{B}^2$ in general, unless $\mathsf{A}\mathsf{B}=\mathsf{B}\mathsf{A}$.
3. $e^{\mathsf{A}}$ is invertible and $\left(e^{\mathsf{A}}\right)^{-1}=e^{-\mathsf{A}}\tag{21}$
4. If $t$ is a scalar variable, then $$e^{\mathsf{A}t}=\mathsf{I}+\mathsf{A}t+\dfrac{\mathsf{A}^2}{2!}t^2+\cdots+\dfrac{\mathsf{A}^n}{n!}t^n+\cdots=\sum^{\infty}_{n=0}\dfrac{\mathsf{A}^n}{n!}t^n,\tag{22}$$
5. Differentiating $e^{\mathsf{A}t}$ with respect to $t$, we obtain $$\begin{align}\dfrac{d}{dt}e^{\mathsf{A}t}&=\mathsf{A}+\mathsf{A}^2t+\cdots+\dfrac{\mathsf{A}^n}{(n-1)!}t^{n-1}+\cdots=\sum^{\infty}_{n=1}\dfrac{\mathsf{A}^n}{(n-1)!}t^{n-1}\\&=\mathsf{A}\underbrace{\left(\mathsf{I}+\mathsf{A}t+\cdots+\dfrac{\mathsf{A}^{n-1}}{(n-1)!}t^{n-1}+\cdots\right)}_{e^{\mathsf{A}t}}=\mathsf{A}\,\underbrace{\sum^{\infty}_{n=1}\dfrac{\mathsf{A}^{n-1}}{(n-1)!}t^{n-1}}_{e^{\mathsf{A}t}}\\&=\mathsf{A}e^{\mathsf{A}t},\end{align}$$ so we can say:
- $e^{\mathsf{A}t}$ is a solution of $\mathsf{X}'=\mathsf{A}\mathsf{X}$, or
- ==$e^{\mathsf{A}t}$ is a fundamental matrix of $\mathbf{x}'=\mathsf{A}\mathbf{x}$== if $\mathbf{x}$ is any column of $\mathsf{X}$.
6. $e^{\mathsf{A}t}(0)=e^{\mathsf{O}}=\mathsf{I}$.
### Example 3
- :film_frames: [202004015_4.MTS](https://reurl.cc/d0LLvz) 08:14
- p.413, Example 3 of Sec. 5.7
::: warning
Given $\mathsf{A}=\begin{pmatrix}0&3&4\\0&0&6\\0&0&0\end{pmatrix}$, calculate $e^{\mathsf{A}t}$.
:::
#### Solution
$\mathsf{A}^2=\begin{pmatrix}0&0&18\\0&0&0\\0&0&0\end{pmatrix}$
$\mathsf{A}^3=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}$
$\mathsf{A}^4=\mathsf{A}^5=\cdots=\mathsf{O}_{3\times 3}$
$\begin{align}e^{\mathsf{A}t}&=\mathsf{I}+\mathsf{A}t+\frac{1}{2!}\mathsf{A}^2t^2\\&=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+\begin{pmatrix}0&3t&4t\\0&0&6t\\0&0&0\end{pmatrix}+\begin{pmatrix}0&0&9t^2\\0&0&0\\0&0&0\end{pmatrix}\\&=\boxed{\begin{pmatrix}1&3t&4t+9t^2\\0&1&6t\\0&0&1\end{pmatrix}}.\;\blacksquare\end{align}$
If $\mathsf{A}^k=\mathsf{O}$ for some positive integer $n$, then $\mathsf{A}$ is said to be **nilpotent**. If $\mathsf{A}$ is nilpotent, then $e^{\mathsf{A}}$ or $e^{\mathsf{A}t}$ has finite terms.
### Example 4
- :film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 00:47
- p.413, Example 4 of Sec. 5.7
::: warning
Given $\mathsf{A}=\begin{pmatrix}2&3&4\\0&2&6\\0&0&2\end{pmatrix}$, calculate $e^{\mathsf{A}t}$.
:::
#### Solution
$\mathsf{A}=\begin{pmatrix}2&3&4\\0&2&6\\0&0&2\end{pmatrix}=\underbrace{\begin{pmatrix}2&0&0\\0&2&0\\0&0&2\end{pmatrix}}_\mathsf{D}+\underbrace{\begin{pmatrix}0&3&4\\0&0&6\\0&0&0\end{pmatrix}}_\mathsf{B},$ where $\mathsf{D}=2\mathsf{I}$ is a diagonal matrix and $\mathsf{B}$ is a diagonal matrix (seen in the previous example).
Since $\mathsf{D}\mathsf{B}=\mathsf{B}\mathsf{D}$ holds, we have $$\begin{align}e^{\mathsf{A}t}&=e^{\mathsf{D}t+\mathsf{B}t}=e^{\mathsf{D}t}e^{\mathsf{B}t}\\ &=\begin{pmatrix}e^{2t}&0&0\\0&e^{2t}&0\\0&0&e^{2t}\end{pmatrix}\begin{pmatrix}1&3t&4t+9t^2\\0&1&6t\\0&0&1\end{pmatrix}\\&=\boxed{\begin{pmatrix}e^{2t}&3te^{2t}&\left(4t+9t^2\right)e^{2t}\\0&e^{2t}&6te^{2t}\\0&0&e^{2t}\end{pmatrix}}.\;\blacksquare\end{align}$$
### 5) Computation of $e^{\mathsf{A}t}$
:film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 05:10
Since $e^{\mathsf{A}t}$ is a fundamental matrix of $\mathbf{x}'=\mathsf{A}\mathbf{x}$, for I.V.P. (13) $$\mathbf{x}'=\mathsf{A}\mathbf{x},\quad \mathbf{x}(0)=\mathbf{x}_0,$$ the solution if given by $$\mathbf{x}(t)=e^{\mathsf{A}t}\underbrace{\left(e^{\mathsf{A}\cdot 0}\right)^{-1}}_{\mathsf{I}}\mathbf{x}_0=e^{\mathsf{A}t}\mathbf{x}_0.\tag{23}$$
If $e^{\mathsf{A}t}$ is *another* fundamental matrix of $\mathbf{x}'=\mathsf{A}\mathbf{x}$, the solution of I.V.P. (13) is also given by Expression (15): $$\mathbf{x}(t)=\mathsf{\Phi}(t)\mathsf{\Phi}(0)^{-1}\mathbf{x}_0.$$ Equating Expressions (15) and (23), we have $$e^{\mathsf{A}t}\mathbf{x}_0=\mathsf{\Phi}(t)\mathsf{\Phi}(0)^{-1}\mathbf{x}_0$$ or $$e^{\mathsf{A}t}=\mathsf{\Phi}(t)\mathsf{\Phi}(0)^{-1}.\tag{24}$$
### Example 5
- :film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 08:48
- p.414, Example 5 of Sec. 5.7
::: warning
Given $\mathsf{A}=\begin{pmatrix}4&2\\3&-1\end{pmatrix}$, calculate $e^{\mathsf{A}t}$.
:::
#### Solution
We have already found a fundamental matrix of $\mathsf{A}$ in Example 1: $$\mathsf{\Phi}(t)=\begin{pmatrix}2e^{5t}&e^{-2t}\\e^{5t}&-3e^{-2t}\end{pmatrix}$$ and thus $$\mathsf{\Phi}(0)^{-1}=\begin{pmatrix}2&1\\1&-3\end{pmatrix}^{-1}=-\dfrac{1}{7}\begin{pmatrix}-3&-1\\-1&2\end{pmatrix}.$$ Hence Expression (24) gives $$\begin{align}e^{\mathsf{A}t}&=\mathsf{\Phi}(t)\mathsf{\Phi}(0)^{-1}\\&=-\dfrac{1}{7}\begin{pmatrix}2e^{5t}&e^{-2t}\\e^{5t}&-3e^{-2t}\end{pmatrix}\begin{pmatrix}-3&-1\\-1&2\end{pmatrix}\\&=\boxed{\dfrac{1}{7}\begin{pmatrix}6e^{5t}+2e^{-2t}&e^{5t}-2e^{-2t}\\3e^{5t}-3e^{-2t}&e^{5t}+6e^{-2t}\end{pmatrix}}.\;\blacksquare\end{align}$$
### 6) Diagonalization
:film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 12:51
A $n\times n$ square matrix $\mathsf{A}$ is **diagonalizable** if there exist an invertible matrix $\mathsf{T}$ and a diagonal matrix $\mathsf{D}$ such that $$\mathsf{T}^{-1}\mathsf{A}\mathsf{T}=\mathsf{D},\tag{25}$$ (in which case $\mathsf{A}$ is said to be **similar** to $\mathsf{D}$.)
Substituting $\mathsf{A}=\mathsf{T}\mathsf{D}\mathsf{T}^{-1}$ into $e^{\mathsf{A}t}=\sum^{\infty}_{n=0}\left(\mathsf{A}^nt^n/n!\right)$, we obtain $$\require{cancel}\begin{align}e^{\mathsf{A}t}&=\sum^{\infty}_{n=0}\left[\dfrac{t^n}{n!}\left(\mathsf{T}\mathsf{D}\mathsf{T}^{-1}\right)^n\right]\\&=\sum^{\infty}_{n=0}\left[\dfrac{t^n}{n!}\overbrace{\left(\mathsf{T}\mathsf{D}\cancel{\mathsf{T}^{-1}}\right)\left(\cancel{\mathsf{T}}\mathsf{D}\cancel{\mathsf{T}^{-1}}\right)\cdots\left(\cancel{\mathsf{T}}\mathsf{D}\mathsf{T}^{-1}\right)}^{n\text{ terms}}\right]\\&=\sum^{\infty}_{n=0}\left(\dfrac{t^n}{n!}\mathsf{T}\mathsf{D}^n\mathsf{T}^{-1}\right)=\mathsf{T}\sum^{\infty}_{n=0}\left(\dfrac{t^n}{n!}\mathsf{D}^n\mathsf{T}^{-1}\right)=\mathsf{T}\sum^{\infty}_{n=0}\left(\dfrac{t^n}{n!}\mathsf{D}^n\right)\mathsf{T}^{-1}\\&=\mathsf{T}e^{\mathsf{D}t}\mathsf{T}^{-1}.\end{align}\tag{26}$$
If we replace $\mathsf{A}t$ by $\mathsf{A}$ in (26), the relation $$e^{\mathsf{A}}=\mathsf{T}e^{\mathsf{D}}\mathsf{T}^{-1}\tag{27}$$ provides another way to calculate $e^{\mathsf{A}}$.
Additionally, if matrix $\mathsf{A}$ is diagonalizable, then System (1) $\mathbf{x}'=\mathsf{A}\mathbf{x}$ can be solved by using the relation $\mathsf{A}=\mathsf{T}\mathsf{D}\mathsf{T}^{-1}$ to obtain $$\mathbf{x}'=\mathsf{T}\mathsf{D}\mathsf{T}^{-1}\mathbf{x}$$ or by left-multiplication of $\mathsf{T}^{-1}$, $$\mathsf{T}^{-1}\mathbf{x}'=\mathsf{D}\mathsf{T}^{-1}\mathbf{x}.$$ Define $\mathbf{y}=\mathsf{T}^{-1}\mathbf{x}$ and it can be shown that $\mathbf{y}$ is differentiable and that $\mathbf{y}'=\mathsf{T}^{-1}\mathbf{x}$. Hence, the System (1) can be written as $$\mathbf{y}'=\mathsf{D}\mathbf{y},\tag{28}$$ which is easy to solve because $\mathsf{D}$ is a diagonal matrix. Setting $\mathbf{y}(t)=\begin{pmatrix}y_1(t)&y_2(t)&\cdots&y_n(t)\end{pmatrix}^T$, then System (28) becomes $$\begin{pmatrix}y_1'(t)\\y_2'(t)\\\vdots\\y_n'(t)\end{pmatrix}=\begin{pmatrix}d_1&&&\\&d_2&&\\&&\ddots&\\&&&d_n\end{pmatrix}\begin{pmatrix}y_1(t)\\y_2(t)\\\vdots\\y_n(t)\end{pmatrix}=\begin{pmatrix}d_1y_1(t)\\d_2y_2(t)\\\vdots\\d_ny_n(t)\end{pmatrix},$$ of which solution is $$\begin{pmatrix}y_1(t)\\y_2(t)\\\vdots\\y_n(t)\end{pmatrix}=\begin{pmatrix}c_1e^{d_1t}\\c_2e^{d_2t}\\\vdots\\c_ne^{d_nt}\end{pmatrix}=\begin{pmatrix}e^{d_1}&&&\\&e^{d_2}&&\\&&\ddots&\\&&&e^{d_n}\end{pmatrix}\begin{pmatrix}c_1\\c_2\\\vdots\\c_n\end{pmatrix},$$ where $c_1,c_2,\ldots,c_n$ are arbitrary constants. Thus, system (28) has a general solution $$\mathbf{y}(t)=e^{\mathsf{D}t}\mathbf{c},\tag{29}$$ where $\mathbf{c}=\begin{pmatrix}c_1&c_2&\cdots&c_n\end{pmatrix}^T$. Finally, a general solution of System (1) is $$\mathbf{x}(t)=\mathsf{T}\mathbf{y}(t)=\mathsf{T}e^{\mathsf{D}t}\mathbf{c}.\tag{30}$$
To illustrate the method of diagonalization, let's see an example.
### Example 6
:film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 08:13
::: warning
Given the system $$\mathbf{x}'=\underbrace{\begin{pmatrix}1&1\\4&1\end{pmatrix}}_{\mathsf{A}}\mathbf{x},\tag{31}$$ find its general solution by diagonizing $\mathsf{A}$.
:::
#### Solution
Solving characterisitc equation of $\mathsf{A}$, $$\begin{vmatrix}1-\lambda&1\\4&1-\lambda\end{vmatrix}=\lambda^2-2\lambda-3=(\lambda-3)(\lambda+1)=0,$$ we obtain $\lambda_1=3$, $\lambda_2=-1$.
For $\lambda_1=3$, the equation for $\mathbf{v}_1=\begin{pmatrix}a&b\end{pmatrix}^T$ $$\begin{pmatrix}-2&1\\4&-2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ yields $\mathbf{v}_1=\begin{pmatrix}1&2\end{pmatrix}^T$.
For $\lambda_2=-1$, the equation for $\mathbf{v}_1=\begin{pmatrix}a&b\end{pmatrix}^T$ $$\begin{pmatrix}2&1\\4&2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ yields $\mathbf{v}_1=\begin{pmatrix}1&-2\end{pmatrix}^T$.
If we define $$\mathsf{T}=\begin{pmatrix}\mathbf{v}_1&\mathbf{v}_2\end{pmatrix}=\begin{pmatrix}1&1\\2&-2\end{pmatrix}\quad\text{and}\quad\mathsf{D}=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}=\begin{pmatrix}3&0\\0&-1\end{pmatrix},$$ then we can check $\mathsf{T}^{-1}\mathsf{A}\mathsf{T}=\mathsf{D}$. Let $\mathbf{y}(t)=\mathsf{T}^{-1}\mathbf{x}(t)$ and use (29) to get $$\mathbf{y}(t)=\begin{pmatrix}e^{3t}&0\\0&e^{-t}\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}=\begin{pmatrix}c_1e^{3t}\\c_2e^{-t}\end{pmatrix}.$$ Finally, using (30),a general solution of the asked system is $$\mathbf{x}(t)=\mathsf{T}\mathbf{y}(t)=\begin{pmatrix}1&1\\2&-2\end{pmatrix}\begin{pmatrix}c_1e^{3t}\\c_2e^{-t}\end{pmatrix}=\boxed{\begin{pmatrix}c_1e^{3t}+c_2e^{-t}\\2c_1e^{3t}-2c_2e^{-t}\end{pmatrix}}.\;\blacksquare\tag{32}$$
### Example 7
:film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 26:45
::: warning
Solve the linear system $$\mathbf{x}''=\underbrace{\begin{pmatrix}-3&2\\2&-3\end{pmatrix}}_{\mathsf{A}}\mathbf{x}.\tag{33}$$
:::
#### Solution
We wish to diagonalize $\mathsf{A}$.
Starting with $$\begin{vmatrix}-3-\lambda&2\\2&-3-\lambda\end{vmatrix}=\lambda^2+6\lambda+5=(\lambda+1)(\lambda+5)=0,$$ we find the eigenvalues of $\mathsf{A}$ are $\lambda_1=-1$, $\lambda_2=-5$.
**Case I** $\lambda_1=-1$: The equation $(\mathsf{A}-\lambda_1\mathsf{I})\mathbf{v}_1=\mathbf{0}$ for $\mathbf{v}_1=\begin{pmatrix}a&b\end{pmatrix}^T$ $$\begin{pmatrix}-2&2\\2&-2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ yields $\mathbf{v}_1=\begin{pmatrix}1&1\end{pmatrix}^T$ associated with $\lambda_1=-1$.
:film_frames: [202004015_6.MTS](https://reurl.cc/Nj4l85) 00:00
**Case II** $\lambda_1=-5$: The equation $(\mathsf{A}-\lambda_2\mathsf{I})\mathbf{v}_2=\mathbf{0}$ for $\mathbf{v}_2=\begin{pmatrix}a&b\end{pmatrix}^T$ $$\begin{pmatrix}2&2\\2&2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ yields $\mathbf{v}_2=\begin{pmatrix}1&-1\end{pmatrix}^T$ associated with $\lambda_2=-5$.
Letting $\mathsf{T}=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ and $\mathsf{D}=\begin{pmatrix}-1&0\\0&-5\end{pmatrix}$, we have $\mathsf{T}^{-1}\mathsf{A}\mathsf{T}=\mathsf{D}$. Note that if we had **normalized** the eigenvectors when determining them, $\mathsf{T}$ would be an **orthogonal** matrix, of which inverse is just its transpose: $\mathsf{T}^{-1}=\mathsf{T}^T$.
Turn back to System (33) and rewrite it as $$\mathbf{x}''=\mathsf{T}\mathsf{D}\mathsf{T}^{-1}\mathbf{x}$$ or $$\mathsf{T}^{-1}\mathbf{x}''=\mathsf{D}\mathsf{T}^{-1}\mathbf{x},$$ or $$\mathbf{y}''=\mathsf{D}\mathbf{y}\tag{34}$$ if we let $\mathbf{y}(t)=\mathsf{T}^{-1}\mathbf{x}(t)=\begin{pmatrix}y_1(t)&y_2(t)\end{pmatrix}^T.$ Equation (34) corresponds to the equations $$\left\{\begin{array}{l}y_1''(t)=-y_1(t)\\y_2''(t)=-5y_2(t)\end{array}\right.,$$ of which the general solutions are $$\left\{\begin{array}{l}y_1(t)=c_1\sin t+c_2\cos t\\y_2(t)=c_3\sin \sqrt{5}t+c_4\cos\sqrt{5} t\end{array}\right..$$
Finally, the general solution of System (33) is $$\begin{align}\mathbf{x}(t)&=\mathsf{T}\mathbf{y}(t)=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}c_1\sin t+c_2\cos t\\c_3\sin \sqrt{5}t+c_4\cos\sqrt{5} t\end{pmatrix}\\&=\boxed{\begin{pmatrix}c_1\sin t+c_2\cos t+c_3\sin \sqrt{5}t+c_4\cos\sqrt{5} t\\c_1\sin t+c_2\cos t-c_3\sin \sqrt{5}t-c_4\cos\sqrt{5} t\end{pmatrix}}.\;\blacksquare\end{align}$$
### 7) Summary
:film_frames: [202004015_5.MTS](https://reurl.cc/qdrrxN) 11:51-12:50
1. If $\mathsf{D}=\text{diag}\left(d_1,d_2,\ldots,d_n\right)$ is a **diagonal** matrix, then $e^\mathsf{D}=\text{diag}\left(e^{d_1},e^{d_2},\ldots,e^{d_n}\right)$.
2. If $\mathsf{B}$ is a **nilpotent** matrix such that $\mathsf{B}^k=\mathsf{O}$ for some $k\in\mathbb{N}$, then $e^\mathsf{B}=\sum_{n=0}^{k-1}\left(\mathsf{B}^n/n!\right)$.
3. If $\mathsf{\Phi}$ is a fundamental matrix of the homogeneous system $\mathbf{x}'=\mathsf{A}\mathbf{x}$, then $e^{\mathsf{A}t}=\mathsf{\Phi}(t)\mathsf{\Phi}(0)^{-1}$.
4. If there exist an invertible matrix $\mathsf{T}$ and a diagonal matrix $\mathsf{D}$ such that $\mathsf{T}^{-1}\mathsf{A}\mathsf{T}=\mathsf{D}$, then $e^{\mathsf{A}}=\mathsf{T}e^{\mathsf{D}}\mathsf{T}^{-1}$.
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