MIT18.06 Linear Algebra (24)

lec24

Markov Matrix

$$A=\left[\begin{array}{ccc}.1 & .01 & .3 \\.2 & .99 & .3 \\.7 & 0 & .4\end{array}\right]$$

properties

1. All entryis $\ge 0$
2. All columns add to $1$
3. $\lambda =1$ is an eigen value
4. all other eigen value $|\lambda_i|<1$

steady state

$$u_k=A^ku_0=c_1\lambda _1x_1+c_2\lambda _2x_2+\cdots+c_n\lambda _nx_n$$

if $\lambda_1=1$ , the steady state of $u_k$ ( $k\to \infty$ ) is $c_1x_1$ i.e. the $x_1$ part of $u_0$ .

why $\lambda =1$ is an eigenvalue?

$$A-1 I=\left[\begin{array}{ccc}-.9 & .01 & .3 \\.2 & -.01 & .3 \\.7 & 0 & -.6\end{array}\right]$$

All columns in $A-I$ adds up to $0$ $\stackrel{?}{\longrightarrow} $ $ A-I$ is singular

Beacuse rows are dependent.

$$row_1+row_2+\cdots+row_n=0$$

i.e. $(1,1,1)$ is in $N((A-I)^\top)$

Application of Marcov matrix: population

$$\begin{bmatrix}u_{cal} \\u_{mass}\end{bmatrix}_{0}=\begin{bmatrix}0\\1000\end{bmatrix}$$

$$\begin{bmatrix}u_{cal} \\u_{mass}\end{bmatrix}_{k+1} =\begin{bmatrix}.9 & .2 \\.1 & .8\end{bmatrix}\begin{bmatrix}u_{cal} \\u_{mass}\end{bmatrix}_{k}$$

$.1 $of California population$ \to$ Massachusetts

$.2 $of Massachusetts population$ \to$ California

find eigenvalue and eigenvectors:

$$\lambda = 1, .7$$

$$u_0=\begin{bmatrix}0 \\ 1000\end{bmatrix}=\frac{1000}{3} \begin{bmatrix}2 \\ 1\end{bmatrix}+\frac{2000}{3} \begin{bmatrix}-1 \\ 1\end{bmatrix}$$

$$u_{k}=c_{1}{\color{Red} 1} ^{k}\begin{bmatrix} 2 \\ 1 \end{bmatrix}+c_{2}(.7)^{k}\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$

steady state

$$u_k=\frac{1000}{3} \begin{bmatrix}2 \\ 1\end{bmatrix}$$

Fourier Series

intro

$x_i$ are orthonormal basis.

$$v=x_1q_1+x_2q_2+\dots+x_nq_n$$

How to get $x_i$ ?

e.g.

$$q_1^\top v=x_1q_1q_1^\top+0+\cdots+0=x_1$$

i.e. (writing in the form of matrix)

$$\begin{aligned} \begin{bmatrix} \vdots & & \vdots\\ q_1 & \cdots & q_n\\ \vdots & & \vdots \end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} & = v \\ Qx & = v \end{aligned}$$

$$x=Q^{-1}v=Q^\top v$$

Fourier Series

$$f(x)=a_{0}+a_{1} \cos x+b_{1} \sin x+a_{2} \cos 2 x+b_{2} \sin 2 x+\cdots$$

$q_i $are orthogonal, and$ \sin $,$ \cos$ are orthogonal funtions. How come?

orthogonal funtions

like inner product, $f=\cos x$ and $g=\sin x$ are orthogonal funtions means:

$$f^{\top} g=\int_{0}^{2 \pi} f(x) g(x) d x=\left.\frac{1}{2}(\sin x)^{2}\right|_{0}^{2 \pi}=0$$

find the coefficient:

$$a_1=\frac{f^\top \cos(x)}{\cos^\top \cos(x)}=\frac{\int_{0}^{2\pi}f(x)\cos(x) \mathrm{d}x}{\int_{0}^{2\pi}\cos(x)^2\mathrm{d}x}=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\cos(x) \mathrm{d}x$$