Diagonalization, power of a

One 、 Diagonalization of matrix and its conditions

1.2 Diagonalization of matrix

The square array $A$ The eigenvector of is put into a new matrix $S$, So there are ：
$$\begin{array}{rl}AS& =A\left[\begin{array}{cccc}{c}_1& c_2& \cdots & c_n\end{array}\right]\\ & =\left[\begin{array}{ccc}{\lambda }_1{c}_1& \cdots & {\lambda }_n{c}_n\end{array}\right]\\ & =\left[\begin{array}{cccc}{c}_1& c_2& \cdots & c_n\end{array}\right]\left[\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots & {\lambda }_n\end{array}\right]=A\Lambda \end{array}\text{\hspace{1em}(1)}$$
There is a ：
$$AS=S\Lambda \text{\hspace{1em}(2)}$$
Because eigenvectors are linearly independent , therefore ：
$$S^{-1}AS=\Lambda \text{\hspace{1em}(3)}$$
The following is the equivalent expression , That is the important formula of matrix diagonalization in our lesson ：
$$A=S\Lambda S^{-1}\text{\hspace{1em}(4)}$$
This is another method of matrix decomposition ： A matrix is divided into a matrix composed of eigenvalue vectors and a matrix composed of eigenvalues . Such decomposition , It makes it feasible for us to solve the power of the matrix .
from （3） There are the following conclusions ：
$$\begin{gathered} Ax=\lambda x \\ {A}^{2}x=\lambda Ax={\lambda }^{2}x \\ \cdots \\ {A}^{n}x={\lambda }^{n}x\text{\hspace{1em}(5)} \end{gathered}$$
from （4） Yes ：
$$\begin{gathered} A^2=S\Lambda S^{-1}S\Lambda S^{-1}=S{\Lambda }^2{S}^{-1} \\ {A}^k=S{\Lambda }^k{S}^{-1}\text{\hspace{1em}(6)} \end{gathered}$$
The existence of eigenvalues gives us an expression for calculating the power of a matrix , Be careful Matrix diagonalization condition yes ： There is $n$ Two independent eigenvectors .

Theorem ： hypothesis $A$ There is $n$ Two independent eigenvectors , When $k\to \infty$,$A^k\to 0$ Is the condition of ：
$$\forall |{\lambda }_i|<1\text{\hspace{1em}(7)}$$
Such a matrix $A$ It is stable. .

Since there are so many good properties of diagonalization , So what kind of matrix can be diagonalized ？ Here is the conclusion ：

$A$ There must be $n$ Two independent vectors （ Or it can be diagonalized （diagnalizable））, If the following conditions are met ：

• all $\lambda$ Are different （ That is, there is no repetition $\lambda$, The textbook has proof ）

Whether repeated eigenvalues must not be diagonalized ？ answer ： Look at the situation . The following is the two clock situation .

Example 1： Unit matrix $I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$, There are three eigenvalues , All are 1, But all vectors are their eigenvectors , Take whatever you like 3 Just one .

Example 2： For a triangular matrix $A=\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]$ The eigenvalue of is 2 and 2, Then find its eigenvector $A-2I=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, The zero space of this matrix is a zero vector , The eigenvector is $x_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$, Because I can't find 2 Linear independent vectors , No diagonalization .

Two 、 Application of diagonalization

If $x_i$ yes $A$ An eigenvector , that $k{x}_i$($k\ne 0$) It's also its eigenvector .

A vector $u_0$ It's a matrix $A$ Eigenvector $x_i$ The linear combination of ：
$$u_0=c_1{x}_1+c_2{x}_2+\cdots +c_n{x}_n\text{\hspace{1em}(8)}$$
Matrix form ：
$$u_0=Sc\text{\hspace{1em}(9)}$$
For everyone after $u_k$ Through the general term ：
$$u_{k+1}=A{u}_k\text{\hspace{1em}(10)}$$
Find out , Concrete ：

$$\begin{gathered} u_1=A{u}_0 \\ {u}_2=A^2{u}_0 \\ \cdots \\ {u}_k=A^k{u}_0\text{\hspace{1em}(10)} \end{gathered}$$

$$\begin{gathered} u_1=A{u}_0=c_1{\lambda }_1{x}_1+c_2{\lambda }_2{x}_2+\cdots +c_n{\lambda }_n{x}_n \\ {u}_{100}=A^{100}{u}_0=c_1{\lambda }_1^{100}{x}_1+c_2{\lambda }_2^{100}{x}_2+\cdots +c_n{\lambda }_n^{100}{x}_n\text{\hspace{1em}(11)} \end{gathered}$$
In matrix form ：
$$\begin{gathered} u_0=Sc \\ {u}_{100}=u_0{A}^{100}={\Lambda }^{100}Sc\text{\hspace{1em}(12)} \end{gathered}$$
Here will be the initial vector $u_0$ Expand into a linear combination of several eigenvectors , So we can deduce $u_{100}$ Result .

Fibonacci sequence （Fibonacci） Example ：
$$0,1,1,2,3,5,8,13\cdots F_{100}$$
How to put the first 100 How to calculate Fibonacci number and its growth rate ？

answer ： Growth rate 、 The convergence 、 Stability is determined by eigenvalues , The recurrence formula is given below ：
$$F_{k+2}=F_{k+1}+F_k\text{\hspace{1em}(13)}$$
This equation is not only related to the previous value $F_k$ of , And also with the value of the previous $F_k$ of , It is a second-order difference equation .(13) You can also add a condition ：
$$\begin{gathered} F_{k+2}=F_{k+1}+F_k \\ {F}_{k+1}=F_{k+1}\text{\hspace{1em}(14)} \end{gathered}$$
In matrix form ：
$$\left[\begin{array}{c}{F}_{k+2}\\ F_{k+1}\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}{F}_{k+1}\\ F_k\end{array}\right]$$
Here we use a little trick to transform the second-order equation “ drop ” Is a first-order equation ：
$$\begin{gathered} u_k=\left[\begin{array}{c}{F}_{k+1}\\ F_k\end{array}\right] \\ {u}_{k+1}=\left[\begin{array}{c}{u}_{k+2}\\ u_{k+1}\end{array}\right]\text{\hspace{1em}(15)} \end{gathered}$$
Plug in （14） Yes ：
$$u_{k+1}=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]u_k\text{\hspace{1em}(16)}$$
from ：$$|A-\lambda I|=-{\lambda }^2-\lambda +1=0$$ The eigenvalues can be obtained as ：
$$\begin{gathered} {\lambda }_1=\frac{1+\sqrt{5}}{2}\approx 1.618 \\ {\lambda }_2=\frac{1-\sqrt{5}}{2}\approx 0.618\text{\hspace{1em}(17)} \end{gathered}$$
The corresponding eigenvector is ：
$$\left[\begin{array}{cc}1-\lambda & 1\\ 1& -\lambda \end{array}\right]x=0\text{\hspace{1em}(18)}$$
$x=\left[\begin{array}{c}\lambda \\ 1\end{array}\right]$ yes （18） Eigenvector of , This is because ：
$$Ax=\left[\begin{array}{c}-{\lambda }^2-\lambda +1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\text{\hspace{1em}(19)}$$
This is because $-{\lambda }^2-\lambda +1=0$ Exactly equal to the characteristic root equation .

$$x_1=\left[\begin{array}{c}1.618\\ 1\end{array}\right]x_2=\left[\begin{array}{c}0.618\\ 1\end{array}\right]\text{\hspace{1em}(20)}$$

In order to take advantage of the previous conclusion , We need to set the initial value $u_0$ use $A$ Linear combination representation of eigenvectors of matrix , Suppose this linear combination is $c_0$ and $c_1$
$$u_0=c_1{x}_1+c_0{x}_2\text{\hspace{1em}(21)}$$
It doesn't matter what the result is . Let's review the whole problem-solving logic ：

• Convert the second-order difference equation into the first-order difference equation , The purpose is to apply the power characteristics of the first-order difference equation
• Determine that the obtained eigenvalues and eigenvectors can be used to represent $u_0$
• If the second step exists , So there are $u_k=A^{100}{u}_0=c_1{\lambda }_1^{100}{x}_1+c_2{\lambda }_2^{100}{x}_2+\cdots +c_n{\lambda }_n^{100}{x}_n$

Because this is only a second-order matrix , therefore ：
$$u_k=c_1{\lambda }_1^{100}{x}_1+c_2{\lambda }_2^{100}{x}_2\text{\hspace{1em}(22)}$$

For this example ：
$$u_{100}=c_1(0.618{)}^{100}+c_2(1.618{)}^{100}\text{\hspace{1em}(23)}$$
Simple analysis , As the order increases , The eigenvalue is less than 1 Of （0.618） The impact will be smaller and smaller , So the eigenvalue of the decisive factor is the larger one （1.618）.