# L06
###### tags: `微分方程特論`
###### date: `8 Apr 2020`
（第2次線上上課）
# Review of Lesson 5
Sec. 5.4: The Eigenvalue Method for Homogeneous Systems
---
:film_frames: [20200408_1.MTS](https://reurl.cc/pd5kpa) starts form 00:00

A linear system of equations with constant coefficients can be written as $$\dfrac{d\mathbf{x}}{dt}=\mathsf{A}\mathbf{x},\tag{1}$$where $\mathsf{A}$ is an $n\times n$ matrix with constant coefficients (i.e., independnent of $t$).

For such a system, we consider a solution of form$$\mathbf{x}=\mathbf{v}e^{\lambda t},$$ and substitute it into equation (1):$$\lambda\mathbf{v}e^{\lambda t}=\mathsf{A}\mathbf{v}e^{\lambda t}.\tag{2}$$
There follows $$\lambda\mathbf{v}=\mathsf{A}\mathbf{v}\tag{3}$$ or $$(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}=\mathbf{0},\tag{4}$$ where $\mathsf{I}$ is the $n\times n$ identity matrix.

:film_frames: [20200408_1.MTS](https://reurl.cc/pd5kpa) 2:15
Equation (4) can have nontrivial solutions (i.e. solutions other than $\mathbf{v}=\mathbf{0}$) only if$$\det(\mathsf{A}-\lambda\mathsf{I})=0.\tag{5}$$

We can hereby confirm that the scalar $\lambda$ is an eigenvalue of $\mathsf{A}$, and the vector $\mathbf{v}$ is the eigenvector of $\mathsf{A}$ that is associated with $\lambda$.

Since $\mathsf{A}$ is an $n\times n$ matrix, it has $n$ eigenvalues, each of which may or may not differ from the others. Therefore we need the following discussion.

:film_frames: [20200408_1.MTS](https://reurl.cc/pd5kpa) 3:55
### Case 1: distinct real eigenvalues
Suppose that Equation (5) has $n$ distinct real roots $\lambda_i(i=1,2,\ldots,n)$ and eigenvectors $\mathbf{v}_i(i=1,2,\ldots,n)$, where $\mathbf{v}_i$ corresponds to $\lambda_i$. According to theory of linear algebra, $\{\mathbf{v}_i\}^n_{i=1}$ is a set of linearly independent vectors.
Also, we have the following results:
- $\{\mathbf{v}_ie^{\lambda_i t}\}^n_{i=1}$ is also a set of linearly independent functions, each of which satisfies Equation (1).
- We can prove this by using the Wronskian of $\mathbf{v}_ie^{\lambda_i t}$.
- $\{\mathbf{v}_ie^{\lambda_i t}\}^n_{i=1}$ is called a **fundamental solution set** of Equation (1).
- The **general solution** of Equation (1) is $$\displaystyle\mathbf{x}(t)=\sum^n_{i=1}c_i\mathbf{v}_ie^{\lambda_i t},\tag{6}$$where $c_i$ are arbitrary constants

# Lesson 06
Sec. 5.4: The Eigenvalue Method for Homogeneous Systems
---
### Example of Case 1
- :film_frames: [20200408_1.MTS](https://reurl.cc/pd5kpa) 7:13
::: warning
Find the general solution of $$\left\{\begin{matrix}\begin{array}{l}\dfrac{dx_1}{dt}=7x_1-x_2+6x_3\\
\dfrac{dx_2}{dt}=-10x_1+4x_2-12x_3\\
\dfrac{dx_3}{dt}=-2x_1+x_2-x_3\end{array}\end{matrix}\right.\tag{7}$$
:::
#### Solution
First we rewrite the system as $$\dfrac{d\mathbf{x}}{dt}=\underbrace{\begin{pmatrix}7&-1&6\\-10&4&-12\\-2&1&-1\end{pmatrix}}_{\text{matrix }\mathbf{A}}\underbrace{\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}}_{\text{vector }\mathbf{x}}.\tag{8}$$

And then we solve the eigenvalues with the secular (characteristic) equation $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}=\mathbf{0}$, or $$\begin{align}0&=\begin{vmatrix}7-\lambda&-1&6\\-10&4-\lambda&-12\\-2&1&-1-\lambda\end{vmatrix}\\&=(7-\lambda)(4-\lambda)(-1-\lambda)-24+60+12(4-\lambda)+12(7-\lambda)+10(1+\lambda)\\&=\lambda^3-10\lambda^2-31\lambda-30\\&=(\lambda-2)(\lambda-3)(\lambda-5)\end{align}$$ The solutions of $\lambda$ are $\lambda=2,3,5.$

::: spoiler
To factorize $\lambda^3-10\lambda^2-31\lambda-30$, we use the [rational root theorem](https://en.wikipedia.org/wiki/Rational_root_theorem) and substitute:
$\lambda=1:\quad 1-10+31-30\neq 0$, so $\lambda-1$ is a factor;
$\lambda=2:\quad 8-40+62-30 = 0$, so $\lambda-2$ is a factor.
$\begin{align}\lambda^3-10\lambda^2-31\lambda-30&=(\lambda-2)(\lambda^2-10\lamdba+15)\\(\lambda-2)(\lambda-3)(\lambda-5)\end{align}$
:::
\
For ==$\lambda_1 =2$==, we can find its associated eigenvector by solving $$\begin{pmatrix}5&-1&6\\-10&2&-12\\-2&1&-3\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$(We can simply expand it, rather than perform Gaussian elimination or row reducing.)

From $$\left\{\begin{array}{r}5a-b+6c&=0\\-10a+2b-12c&=0\\-2a+b-3c&=0\end{array}\right.,$$ we find that:
- the first and the second equations are dependent;
- adding the first and the third equations gives $3a+3c=0$ or $a+c=0$

Thus if we choose $a=1$, then $c=-1$ and substituting these into the first equation gives $b=5a+6c=-1$.

So the eigenvector which associated with $\lambda_1=1$ is ==$\mathbf{v}_1=\begin{pmatrix}1&-1& -1\end{pmatrix}^T$== and the first solution is $$\mathbf{x}_1=\begin{pmatrix}1\\-1\\ -1\end{pmatrix}e^{2t}.$$

For ==$\lambda_2 =3$==, we have the equation $$\begin{pmatrix}4&-1&6\\-10&1&-12\\-2&1&-4\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$ which gives $$\left\{\begin{array}{r}4a-b+6c&=0\\-10a+b-12c&=0\\-2a+b-4c&=0\end{array}\right.,$$ from which:
- adding the first and the second equations we obtain $-6a-6c=0$ or $a+c=0$;
- adding the first and the second equations we obtain $-8a-8c=0$, which is dependent with the previous one.

If we choose $a=1$, then $c=-1$, and $b=4a+6c=-2$.

So the eigenvector associated with $\lambda_2=3$ is ==$\mathbf{v}_2=\begin{pmatrix}1&-2& -1\end{pmatrix}^T$== and the second solution is $$\mathbf{x}_2=\begin{pmatrix}1\\-2\\ -1\end{pmatrix}e^{3t}.$$

For ==$\lambda_3 =5$==, we have the equation $$\begin{pmatrix}2&-1&6\\-10&-1&-12\\-2&1&-6\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$ which gives $$\left\{\begin{array}{r}2a-b+6c&=0\\-10a-b-12c&=0\\-2a+b-6c&=0\end{array}\right.,$$ from which we find:
- adding the first and the second equations gives $12a+18c=0$ or $2a+3c=0$;
- the first and the third equations are dependent.

If we choose $a=3$, then $c=-2$, and $b=2a+6c=-6$.

So the eigenvector associated with $\lambda_3=5$ is ==$\mathbf{v}_3=\begin{pmatrix}3&-6& -2\end{pmatrix}^T$== and the third solution is $$\mathbf{x}_3=\begin{pmatrix}3\\-6\\ -2\end{pmatrix}e^{5t}.$$

Finally, with the three (linearly independent) solutions we've found, we obtain the general solution $$\boxed{\mathbf{x}(t)=c_1\begin{pmatrix}1\\-1\\ 1\end{pmatrix}e^{2t}+c_2\begin{pmatrix}1\\-2\\ -1\end{pmatrix}e^{3t}+c_3\begin{pmatrix}3\\-6\\ -2\end{pmatrix}e^{5t}},\tag{9}$$ where $c_1,c_2,c_3$ are arbitrary real constants. $\blacksquare$

#### MATLAB
**Input**

A = [7 -1 6;-10 4 -12;-2 1 -1];
[V,D] = eig(A)

**Output**

V =

  -0.42857  -0.40825  -0.57735
   0.85714   0.81650   0.57735
   0.28571   0.40825   0.57735

D =

Diagonal Matrix

   5.0000        0        0
        0   3.0000        0
        0        0   2.0000

---
:film_frames: [20200408_1.MTS](https://reurl.cc/pd5kpa) 19:55
### Case 2: complex eigenvalues
For simplicity, suppose that $n=2$ for System (1) so that $\mathsf{A}\in\mathcal{M}_{2\times 2}(\mathbb{R})$. And suppose that the two eigenvalues of $\mathsf{A}$ are **complex conjugate pairs**, say, $\lambda=\alpha+i\beta$ and $\overline{\lambda}=\alpha-i\beta$, where $\alpha$ and $\beta$ are real.

We claim that, if we let $\mathbf{v}$ be the eigenvector associated with $\lambda$, then $\overline{\mathbf{v}}$ will be the eigenvector associated with $\overline{\lambda}$. This can be verified by taking complex conjugate of Equation (4); and then one can see that $\overline{(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}}=\overline{\mathbf{0}}$ leads to $(\mathsf{A}-\overline{\lambda}\mathsf{I})\overline{\mathbf{v}}=\mathbf{0}$.

Now if we substitute $\lambda=\alpha+i\beta$ and $\mathbf{v}=\mathbf{a}+i\mathbf{b}\;(\mathbf{a},\mathbf{b}\in\mathbb{R}^n)$ into the aforementioned form of linearly independent solution $\mathbf{v}e^{\lambda t}$, then a solution of System (1) is found: $$\begin{align}\mathbf{z}_1(t):=\mathbf{v}e^{\lambda t}&=(\mathbf{a}+i\mathbf{b})e^{(\alpha+i\beta)t}\\&=(\mathbf{a}+i\mathbf{b})e^{\alpha t}(\cos{\beta t}+i\sin{\beta t})\\&=e^{\alpha t}\left[(\mathbf{a}\cos\beta t-\mathbf{b}\sin\beta t)+i(\mathbf{b}\cos\beta t+\mathbf{a}\sin\beta t)\right].\end{align}\tag{10a}$$

Since $\overline{\mathbf{v}}$ is another eigenvector, we also have $$\begin{align}\mathbf{z}_2(t):=\overline{\mathbf{v}}e^{\overline{\lambda} t}&=(\mathbf{a}-i\mathbf{b})e^{(\alpha-i\beta)t}\\&=(\mathbf{a}-i\mathbf{b})e^{\alpha t}(\cos{\beta t}-i\sin{\beta t})\\&=e^{\alpha t}\left[(\mathbf{a}\cos\beta t-\mathbf{b}\sin\beta t)-i(\mathbf{b}\cos\beta t+\mathbf{a}\sin\beta t)\right],\end{align}\tag{10b}$$ which is linearly independent on $\mathbf{v}e^{\lambda t}$.

:film_frames:[20200408_2.MTS](https://reurl.cc/xZ6nm5) 00:00

Despite the fact that two linearly independent solutions $\mathbf{z}_1$ and $\mathbf{z}_2$ are found, our ideal solutions are *real-valued* functions, rather than *complex-valued* ones. A proper pair of real-valued linearly independent solutions can be $$\mathbf{x}_1(t):=\dfrac{1}{2}\left(\mathbf{z}_1+\mathbf{z}_2\right)=e^{\alpha t}(\mathbf{a}\cos\beta t-\mathbf{b}\sin\beta t)\tag{11a}$$ and $$\mathbf{x}_2(t):=\dfrac{1}{2}\left(\mathbf{z}_1-\mathbf{z}_2\right)=e^{\alpha t}(\mathbf{b}\cos\beta t+\mathbf{a}\sin\beta t)\tag{11b}$$ Hence, the general solution of System (1) with $n=2$ is $$\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t).$$ It is not necessary to memorize both (11a) and (11b), as long as we recognize that:
- (11a) is the real part of either (10a) or (10b), and that
- (11b) is the imaginary part of either (10a) or (10b).

So the strategy to tackle System (1) in case 2 is to:
1. find one of complex-valued eigenvalues $\lambda$; and to
2. find the complex-valued eigenvector $\mathbf{v}$; and finally to
3. calculate $\Re(\mathbf{v}e^{\lambda t})$ and $\Im(\mathbf{v}e^{\lambda t})$, exactly the real-valued linearly independent solutions.

### Example 1 of Case 2
- :film_frames:[20200408_2.MTS](https://reurl.cc/xZ6nm5) 03:50
- Example 3 of Sec. 5.4, p.375
::: warning

Find a general solution of the system $$\left\{\begin{array}{l}\dfrac{dx_1}{dt}&=4x_1-3x_2\\\dfrac{dx_2}{dt}&=3x_1+4x_2\end{array}.\right.\tag{12}$$
:::
#### Solution
Writing the matrix $$\mathsf{A}=\begin{pmatrix}4&-3\\3&4\end{pmatrix},$$ and solving for eignevalues with $$\begin{align}0&=\begin{vmatrix}4-\lambda&-3\\3&4-\lambda\end{vmatrix}\\&=(\lambda-4)^2+9\\&=\lambda^2-8\lambda+25,\end{align}$$ we find $\lambda=4\pm\sqrt{-9}=4\pm 3i$.

For the eigenvalue $\lambda=4+3i$, the eigenvector is found by the equation $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}=\mathbf{0}$:$$\left\{\begin{array}{rcrcr}(-3i)a&+&(-3)b&=&0\\3a&+&(-3i)b&=&0\end{array}\right.,$$ in which the two equations are dependent. If we chhose $b=i$, then $a=-1$. Thus the eignevector associated with $\lambda=4+3i$ is $\mathbf{v}=\begin{pmatrix}-1&i\end{pmatrix}^T$.

And then we calculate $$\begin{align}\mathbf{v}e^{\lambda t}&=\begin{pmatrix}-1\\i\end{pmatrix}e^{(4+3i)t}=\begin{pmatrix}-1\\i\end{pmatrix}e^{4t}\left(\cos 3t+i\sin 3t \right)\\&=\underbrace{e^{4t}\begin{pmatrix}-\cos 3t\\-\sin 3t\end{pmatrix}}_{\Re(\mathbf{v}e^{\lambda t})}+i\underbrace{e^{4t}\begin{pmatrix}-\sin 3t\\\cos 3t\end{pmatrix}}_{\Im(\mathbf{v}e^{\lambda t})}.\end{align}$$

Therefore, the general solution of System (12) is $$\boxed{\mathbf{x}(t)=c_1e^{4t}\begin{pmatrix}-\cos 3t\\-\sin 3t\end{pmatrix}+c_2e^{4t}\begin{pmatrix}-\sin 3t\\\cos 3t\end{pmatrix}},\tag{13}$$ where $c_1,c_2$ are arbitrary real constants. $\blacksquare$

### Example 2 of Case 2
- :film_frames:[20200408_2.MTS](https://reurl.cc/xZ6nm5) 13:58
- Example 3 of Sec. 5.4, p.376
::: warning
Find a general solution of the system $$\dfrac{d\mathbf{x}}{dt}=\begin{pmatrix}-0.2&0&0.2\\0.2&-0.4&0\\0&0.4&-0.2\end{pmatrix}\mathbf{x}.\tag{14}$$
:::
#### Solution
After solving the secular equation of $\mathsf{A}$: $\det\left(\mathsf{A}-\lambda\mathsf{I}\right)=0$, one can find a real eignevalue $\lambda_1=0$ and a complex conjugate pair $\lambda_2=-0.4- 0.2i$, $\overline{\lambda}_2=-0.4+ 0.2i$.

For $\lambda_1=0$, one can find the associated eigenvector $\mathbf{v}_1=\begin{pmatrix}2&1&2\end{pmatrix}^T$. So the first linearly independnt solution of System (13) is $$\mathbf{x}_1(t)=\begin{pmatrix}2\\1\\2\end{pmatrix}e^{0t}.$$

For $\lambda_2=-0.4- 0.2i$, one can find the associated eigenvector $\mathbf{v}_2=\begin{pmatrix}1&i&-1-i\end{pmatrix}^T$. So the rest linearly independnt solutions of System (13) is the real part and the imaginary part of $$\mathbf{z}_2(t)=\begin{pmatrix}1\\i\\-1-i\end{pmatrix}e^{(-0.4- 0.2i)t}.$$

Therefore, the general solution of System (14) is $$\begin{align}\mathbf{x}(t)&=c_1\mathbf{x}_1(t)+c_2\Re{(\mathbf{z}_2(t))}+c_3\Im{(\mathbf{z}_2(t))}\\&=\boxed{c_1\begin{pmatrix}2\\1\\2\end{pmatrix}+c_2e^{-0.4t}\begin{pmatrix}\cos 0.2t\\ \sin 0.2t\\-\cos 0.2t-\sin 0.2t\end{pmatrix}\\+c_3e^{-0.4t}\begin{pmatrix}-\sin 0.2t\\\cos 0.2t\\-\cos 0.2t+\sin 0.2t\end{pmatrix},}\end{align}\tag{15}$$ where $c_1,c_2,c_3$ are arbitrary real constants. $\blacksquare$

#### MATLAB
**Input**

A=[-0.2 0 0.2;0.2 -0.4 0;0 0.4 -0.2];
[V D]=eig(A)

**Output**

V =

  -0.66667 + 0.00000i   0.35355 + 0.35355i   0.35355 - 0.35355i
  -0.33333 + 0.00000i   0.35355 - 0.35355i   0.35355 + 0.35355i
  -0.66667 + 0.00000i  -0.70711 + 0.00000i  -0.70711 - 0.00000i

D =

Diagonal Matrix

  -0.00000 + 0.00000i                    0                    0
                    0  -0.40000 + 0.20000i                    0
                    0                    0  -0.40000 - 0.20000i

---
Sec. 5.6: Multiple Eigenvalue Solution
---

### Case 3a: "complete" multiple real eigenvalues
:film_frames:[20200408_3.MTS](https://reurl.cc/ar46gY) 00:42
#### Definition
::: success
1. If an eigenvalue $\lambda$ is a $k$-fold root of Equation (3): $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}=\mathbf{0}$, then we call it an eigenvalue of **multiplicity** $k$. (More specifically, it is called **algebraic multiplicity**, in linear algebra.)
2. If the eigenvalue $\lambda$ of multiplicity $k$ has exactly $k$ linearly independent eigenvectors that associate with it, then it is called **complete**; and thus $\left\{\mathbf{v}_ie^{\lambda t}\right\}^k_{i=1}$ are $k$ linearly independent solution of System (1).
:::

### Example of Case 3a
- :film_frames:[20200408_3.MTS](https://reurl.cc/ar46gY) 04:34
- p. 393, Example 1 in Sec. 5.6
:::warning
Find a general solution of $$\mathbf{x}'=\underbrace{\begin{pmatrix}9&4&0\\-6&-1&0\\6&4&3\end{pmatrix}}_\mathsf{A}\mathbf{x}.\tag{16}$$
:::
#### Solution
Writing $(\mathsf{A}-\lambda \mathsf{I})\mathbf{v}=\mathbf{0}$: $$\begin{align}0&=\begin{vmatrix}9-\lambda&4&0\\-6&-1-\lambda&0\\6&4&3-\lambda\end{vmatrix}\\&=(3-\lambda)[(9-\lambda)(-1-\lambda)+24]\\&=(3-\lambda)(\lambda^2-8\lambda+15)\\&=(3-\lambda)^2(5-\lambda),\end{align}$$ we find $\lambda=5,3,3$. Thus we say that $\lambda=3$ is a eigenvalue of multiplicity 2, and that $\lambda=5$ is a eigenvalue of multiplicity 1.

For $\lambda=5$, we solve the eigenvector equation $$\begin{pmatrix}4&4&0\\-6&-6&0\\6&4&-2\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$ and find the associated eigenvector $\mathbf{v}_1=\begin{pmatrix}1&-1&1\end{pmatrix}^T$ and the first linear independent solution of System (16) is $\mathbf{x}_1=\begin{pmatrix}1&-1&1\end{pmatrix}^Te^{5t}$.

For $\lambda=3$, we solve the eigenvector equation $$\begin{pmatrix}6&4&0\\-6&-4&0\\6&4&0\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$ and find that *three equations are dependent, with two variable being underdetermined*.

Hence, if we choose
- $a=1$ and $c=0$, then $b=-\frac{3}{2}$, yieding ==$\mathbf{v}_2=\begin{pmatrix}1&-\frac{3}{2}&0\end{pmatrix}^T$== and the second linearly independent solution $\mathbf{x}_2=\begin{pmatrix}1&-\frac{3}{2}&0\end{pmatrix}^Te^{3t}$;
- $a=0$ and $c=1$, then $b=0$, yieding ==$\mathbf{v}_3=\begin{pmatrix}0&0&1\end{pmatrix}^T$== and the third linearly independent solution $\mathbf{x}_3=\begin{pmatrix}0&0&1\end{pmatrix}^Te^{3t}$.

Therefore, the general solution of System (16) is $$\boxed{\mathbf{x}(t)=c_1\begin{pmatrix}1\\-1\\1\end{pmatrix}e^{5t}+c_2\begin{pmatrix}1\\-\frac{3}{2}\\0\end{pmatrix}e^{3t}+c_3\begin{pmatrix}0\\0\\1\end{pmatrix}e^{3t}},\tag{17}$$ where $c_1,c_2,c_3$ are arbitrary real constants.

*Comment*: Although the multiplicity of $\lambda=3$ is 2, we still can find 2 lineary independent solution from it. Thus, the eigenvalue $\lambda=3$ is complete. $\blacksquare$
#### MATLAB
**Input**

A=[9 4 0;-6 -1 0;6 4 3];
[V D]=eig(A)

**Output**

V =

   0.00000   0.49614  -0.57735
   0.00000  -0.74421   0.57735
   1.00000   0.44721  -0.57735

D =

Diagonal Matrix

   3.0000        0        0
        0   3.0000        0
        0        0   5.0000

### Example of Case 3b
- :film_frames:[20200408_3.MTS](https://reurl.cc/ar46gY) 15:34
- p. 395, Example 2 in Sec. 5.6
::: warning
Find the eigenvectors of the matrix $$\mathsf{A}=\begin{pmatrix}1&-3\\3&7\end{pmatrix}.\tag{18}$$
:::
#### Solution
Solving $$\begin{align}0&=\begin{vmatrix}1-\lambda&-3\\3&7-\lambda\end{vmatrix}\\&=(1-\lambda)(7-\lambda)+9\\&=\lambda^2-8\lambda+16\\&=(\lambda-4)^2,\end{align}$$ we have the eigenvaleu $\lambda=4$ of multiplicity $k=2$.

Solving $$(\mathsf{A}-4\mathsf{I})\mathbf{v}=\begin{pmatrix}-3&-3\\3&3\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix},$$ we only find $\mathbf{v}_1=\begin{pmatrix}1&-1\end{pmatrix}^T$ as the eigenvector associated with $\lambda=4$. $\blacksquare$

### Case 3b: "defective" multiple real eigenvalues
#### Definition
:film_frames:[20200408_3.MTS](https://reurl.cc/ar46gY) 18:08
::: success
3. Let $p$ denote the number of linearly independent eigenvectors associated with the eigenvalue $\lambda$ found by the eigenvalue method. Then,
- if $p = k$, then eigenvalue $\lambda$ is called **complete** (this is consistent to Definition 2);
- if $p < k$, then then eigenvalue $\lambda$ is called **defective**, and the number $$d=k−p$$ of "missing" eigenvectors is called the **defect** of the defective eigenvalue.
:::
For example, $d=3-3=0$ in Example of “complete” multiple eigenvalues, and $d=2-1=1$ in Example of “incomplete” multiple eigenvalues.

Recall the method we used in solving linear homogeneous equaitons. For instance, given $y''-8y'+16y=0$, and knowing that the roots of its characteristic equation are $r=4,4$, we use the linearly independent solutions $e^{4y}$ and $te^{4y}$ to write the general solutions: $$y(t)=c_1e^{4t}+c_2te^{4t}.$$

:film_frames:[20200408_3.MTS](https://reurl.cc/ar46gY) 23:22
Inspiring by this method, we continue on finding another eigenvector associated with a defective eigenvalue.

1. **Trial 1**
Imitating the solution of form $te^{rt}$, we first consider $$\mathbf{x}_2(t)=\mathbf{v}_1te^{\lambda t},\tag{19}$$ where $\mathbf{v}_1$ is a *nonzero* eigenvector of $\mathsf{A}$, associated with $\lambda$, and thus we have $$\mathbf{x}_2'(t)=\mathbf{v}_1e^{\lambda t}+\lambda \mathbf{v}_1te^{\lambda t},$$ such that such that $\mathbf{x}'=\mathsf{A}\mathbf{x}$ becomes $$\mathbf{v}_1e^{\lambda t}+\lambda \mathbf{v}_1te^{\lambda t}=\mathsf{A}\mathbf{v}_1te^{\lambda t},$$ or $$\mathbf{v}_1+\cancel{\lambda \mathbf{v}_1t}=\cancel{\mathsf{A}\mathbf{v}_1t}$$ in which nonvanishing $e^{\lambda t}$ was cancelled. Unfortunately, the above equation has no solution other than $\mathbf{v}_1=\mathbf{0}$ because we can still cancel the terms due to the fact that $\mathsf{A}\mathbf{v}_1=\lambda\mathbf{v}_1$.

2. **Trial 2**
Provided that $\mathbf{v}_1$ is, as mentioned, a *nonzero* eigenvector of $\mathsf{A}$ associated with $\lambda$, now we assume another *nonzero* eigenvector $\mathbf{v}_2$, also associated with $\lambda$, and reconsider the comination $$\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2)e^{\lambda t}=\mathbf{v}_1te^{\lambda t}+\mathbf{v}_2e^{\lambda t},\tag{20}$$, and then we have $$\mathbf{x}_2'(t)=\mathbf{v}_1e^{\lambda t}+\lambda\mathbf{v}_1te^{\lambda t}+\lambda \mathbf{v}_2e^{\lambda t},$$ such that $\mathbf{x}'=\mathsf{A}\mathbf{x}$ becomes $$\require{cancel}\mathbf{v}_1e^{\lambda t}+\lambda\mathbf{v}_1te^{\lambda t}+\lambda \mathbf{v}_2e^{\lambda t}=\mathsf{A}\mathbf{v}_1te^{\lambda t}+\mathsf{A}\mathbf{v}_2e^{\lambda t},$$ which, after collecting terms, yields

:film_frames:[20200408_4.MTS](https://reurl.cc/62QpRk) 00:00$$(\mathbf{v}_1+\lambda \mathbf{v}_2)e^{\lambda t}+\cancel{(\lambda\mathbf{v}_1)te^{\lambda t}}=(\mathsf{A}\mathbf{v}_2)e^{\lambda t}+\cancel{(\mathsf{A}\mathbf{v}_1)te^{\lambda t}},$$ in which the terms of $te^{\lambda t}$ are cancelled due to the fact that $\mathsf{A}\mathbf{v}_1=\lambda\mathbf{v}_1$. After the equation is divided by nonvanishing $e^{\lambda t}$, we are left with $$\mathbf{v}_1+\lambda\mathbf{v}_2=\mathsf{A}\mathbf{v}_2$$ or $$\left(\mathsf{A}-\lambda\mathsf{I}\right)\mathbf{v}_2=\mathbf{v}_1.\tag{21}$$
The function $\mathbf{x}_2=(\mathbf{v}_1t+\mathbf{v}_2)e^{\lambda t}$ is a solution of System (1) if and only if $\mathbf{v}_1$ satisfies Equation (21).

In conclusion, in the case $d=k-p=1$, the two eigenvectors associated with the multiple eigenvalue $\lambda$ are:
1. ==$\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}$==, where $\mathbf{v}_1$ satisfies $$\left(\mathsf{A}-\lambda\mathsf{I}\right)\mathbf{v}_1=\mathbf{0};$$ and
2. ==$\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2) e^{\lambda t}$==, where $\mathbf{v}_2$ satisfies $$\left(\mathsf{A}-\lambda\mathsf{I}\right)\mathbf{v}_2=\mathbf{v}_1$$ or (see Note at :film_frames:[20200408_4.MTS](https://reurl.cc/62QpRk) 09:45-13:11) $$\left(\mathsf{A}-\lambda\mathsf{I}\right)^2\mathbf{v}_2=\mathbf{0}.\tag{22}$$ $\mathbf{v}_2$ is also called a **generalized eigenvector** or an **eigenvector of rank** $\boldsymbol2$ of $\mathsf{A}$.

### Example of Case 3b (Continued)
:film_frames:[20200408_4.MTS](https://reurl.cc/62QpRk) 03:16
::: warning
Given the multiple eigenvalue $\lambda=4$ and its associated eigenvector $\mathbf{v}_1=\begin{pmatrix}1&-1\end{pmatrix}^T$ of the matrix $$\mathsf{A}=\begin{pmatrix}1&-3\\3&7\end{pmatrix}, \tag{18}$$ find the other eigenvector of $\mathsf{A}$.
:::
#### Solution
Let $\mathbf{x}_2(t)=\begin{pmatrix}1&-1\end{pmatrix}^Tte^{4t}+\underbrace{\begin{pmatrix}a&b\end{pmatrix}}_{\mathbf{v}_2}e^{4t}$, where $\mathbf{v}_2$ satisfies $\left(\mathsf{A}-4\mathsf{I}\right)\mathbf{v}_2=\mathbf{v}_1$ or $$\begin{pmatrix}-3&-3\\3&3\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix}.$$ There follows $-3a-3b=1$ and if we choose $a=-\frac{1}{3}$, than $b=0$.

Then we can write the second independent solutions: $$\mathbf{x}_2(t)=\begin{pmatrix}1\\-1\end{pmatrix}te^{4t}+\begin{pmatrix}-\frac{1}{3}\\0\end{pmatrix}e^{4t}=\begin{pmatrix}t-\frac{1}{3}\\-t\end{pmatrix}e^{4t}.$$

Therefore the general solution for System (1) in which $\mathsf{A}$ is given by (18) is $$\begin{align}\mathbf{x}(t)&=c_1\begin{pmatrix}1\\-1\end{pmatrix}e^{4t}+c_2\begin{pmatrix}t-\frac{1}{3}\\-t\end{pmatrix}e^{4t}\\&=\boxed{\begin{pmatrix}e^{4t}\left(c_2t+c_1-\frac{1}{3}c_2\right)\\e^{4t}\left(-c_2t+-c_1\right)\end{pmatrix}}.\blacksquare\end{align}$$

### Summary
:film_frames:[20200408_5.MTS](https://reurl.cc/b5D4RM) 00:00

| 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 |
|:------------:|:------:|:------:| ------------------------------------------------------------------------------------------------------ |
| $k=3$ | $d=2$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^2\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})^3\mathbf{v}_3=\mathbf{0}$ | $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2)e^{\lambda t}\\\color{red}{\mathbf{x}_3(t)=(\frac{1}{2}\mathbf{v}_1t^2+\mathbf{v}_2t+\mathbf{v}_3)e^{\lambda t}}$ |
| $k=3$ | $d=1$ | ? | ?|
| $k=3$ | $d=0$ | $(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_1=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_2=\mathbf{0}\\(\mathsf{A}-\lambda\mathsf{I})\mathbf{v}_3=\mathbf{0}$ |$\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}\\\mathbf{x}_2(t)=\mathbf{v}_2e^{\lambda t}\\\mathbf{x}_3(t)=\mathbf{v}_3e^{\lambda t}$ |

### Example of Case 3b with $k=3$, $d=2$
- :film_frames:[20200408_5.MTS](https://reurl.cc/b5D4RM) 07:21
- p.399, Example 4 of Sec. 5.6
::: warning
Find three linearly independent solutions of the system $$\mathbf{x}'=\underbrace{\begin{pmatrix}0&1&2\\-5&-3&-7\\1&0&0\end{pmatrix}}_\mathsf{A}\mathbf{x}.\tag{23}$$
:::
#### Solution
The secular equation of $\mathsf{A}$ is $$\begin{align}0&=\begin{vmatrix}0-\lambda&1&2\\-5&-3-\lambda&-7\\1&0&-\lambda\end{vmatrix}=-(\lambda+1)^3.\end{align}$$So $\mathsf{A}$ has the eigenvalue $\lambda=-1$ of multiplicity 3.

Solving $(\mathsf{A}+\mathsf{I})\mathbf{v}_1=\mathbf{0}$ or $$\begin{pmatrix}1&1&2\\-5&-2&-7\\1&0&1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$ we find the third row gives $c=-a$ and the first row $a+b+2c=0$ gives $b=a$. So the only assoicated eigenvector is $\mathbf{v}_1=\begin{pmatrix}1&1&-1\end{pmatrix}^T$ and the only linearly independent solution we've found is $$\mathbf{x}_1(t)=\boxed{\begin{pmatrix}1\\1\\-1\end{pmatrix}e^{-t}};\tag{24}$$ this tells us that $p=1$ and dectect $d=2$.

For the second L.I. solution, we consider the equation $(\mathsf{A}+\mathsf{I})\mathbf{v}_2=\mathbf{v}_1$ or $$\begin{pmatrix}1&1&2\\-5&-2&-7\\1&0&1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}1\\1\\-1\end{pmatrix}.$$ This time, we have $\mathbf{v}_2=\begin{pmatrix}-1&2&0\end{pmatrix}^T$ and the second L.I. solution $$\mathbf{x}_2(t)=\boxed{\begin{pmatrix}1\\1\\-1\end{pmatrix}te^{-t}+\begin{pmatrix}-1\\2\\0\end{pmatrix}e^{-t}}.\tag{25}$$

For the third L.I. solution, we consider the equation $(\mathsf{A}+\mathsf{I})\mathbf{v}_3=\mathbf{v}_2$ or $$\begin{pmatrix}1&1&2\\-5&-2&-7\\1&0&1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}-1\\2\\0\end{pmatrix}.$$ We can take $\mathbf{v}_3=\begin{pmatrix}0&-1&0\end{pmatrix}^T$ the third L.I. solution $$\mathbf{x}_3(t)=\boxed{\dfrac{1}{2}\begin{pmatrix}1\\1\\-1\end{pmatrix}t^2e^{-t}+\begin{pmatrix}-1\\2\\0\end{pmatrix}te^{-t}+\begin{pmatrix}0\\-1\\0\end{pmatrix}e^{-t}}.\tag{26}$$

(connection error) :film_frames:[20200408_5.MTS](https://reurl.cc/b5D4RM) 07:21

(after reconnetion) :film_frames:[20200408_6.MTS](https://reurl.cc/QdenA5) 00:00

Therefore, the general solution of System (23) is the combination of solutions (24), (25), and (26): $$\begin{align}\mathbf{x}(t)&=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_3(t)+c_3\mathbf{x}_3(t)\\&=\boxed{c_1\begin{pmatrix}1\\1\\-1\end{pmatrix}e^{-t}+c_2\begin{pmatrix}t-1\\t+2\\-t\end{pmatrix}e^{-t}+c_3\begin{pmatrix}\frac{1}{2}t^2-t\\\frac{1}{2}t^2+2t-1\\-\frac{1}{2}t^2\end{pmatrix}e^{-t}},\end{align}\tag{27}$$ where $c_1,c_2,c_3$ are arbitrary real constants.