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Author: HAO LEE
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Eigen Values and Eigen Vectors

Question 1: How to find Eigenvalues and eigen vectors by hand?

M= [1423]\begin{bmatrix}1&4\\2&3\end{bmatrix}

What are eigenvalues and eigenvectors?

Special combination of a scalar: λ\lambda and a vector vv that:

MV=λVMV = \lambda V

How to find it by hand?

MV=λV    MV=λ[1001]VMV = \lambda V \implies MV = \lambda \begin{bmatrix}1&0\\0&1\end{bmatrix}V

(MλI)V=0    [1λ423λ]V=0(M - \lambda I)V = 0 \implies \begin{bmatrix}1-\lambda&4\\2&3-\lambda\end{bmatrix}V = 0

(1λ)(3λ)4×2=0=λ24λ5=0(1-\lambda)(3-\lambda)-4\times 2 = 0 = \lambda^2-4\lambda-5 = 0

=(λ5)(λ+1)=0    λ=5 or 1 = (\lambda-5)(\lambda+1) =0\implies \lambda = 5 \text{ or } -1 Find Eigen Values

Find Eigen vectors

Let λ=5\lambda = 5

[154235][v1v2]=0    [4422][v1v2]=0 \begin{bmatrix}1-5&4\\2&3-5\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = 0 \implies \begin{bmatrix}-4&4\\2&-2\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = 0

{4v1+4v2=02v12v2=0    v1=v2\begin{cases}-4v_1+4v_2 = 0\\2v_1-2v_2 =0\end{cases}\implies v_1 = v_2 Eigen vector is kR,[kk]//[11]k \in \mathbb{R}, \begin{bmatrix}k\\k\end{bmatrix}//\begin{bmatrix}1\\1\end{bmatrix}

There is always "infinity solutions of eigen vectors!!! (But in the same direction)"

Let λ=1\lambda = -1

[1+1423+1][v1v2]=0    [2424][v1v2]=0 \begin{bmatrix}1+1&4\\2&3+1\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = 0 \implies \begin{bmatrix}2&4\\2&4\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = 0

v1=2k,v2=k, or v=[2kk]kR//[21]v_1 = 2k,v_2 = -k,\text{ or } v = \begin{bmatrix}2k\\-k\end{bmatrix} k\in \mathbb{R}//\begin{bmatrix}2\\-1\end{bmatrix}

Can we diagonalize the matrix?

Let T=[1211]T = \begin{bmatrix}1&2\\1&-1\end{bmatrix}

Find T1T^{-1}!!!

[1211][x1x2]=[b1b2]    [x1+2x2x1x2]=[b1b2]\begin{bmatrix}1&2\\1&-1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}b_1\\b_2\end{bmatrix}\implies\begin{bmatrix}x_1+2x_2\\x_1-x_2\end{bmatrix}=\begin{bmatrix}b1\\b_2\end{bmatrix}

First row + 2×\timesSecond row( I want to cancel x2x_2): 3x1=b1+2b2    x1=b1+2b233x_1 = b_1+2b_2 \implies x_1 = \frac{b_1+2b_2}{3}

x1+2x2=b1    b1+2b23+2x2=b1    2x2=b1b1+2b23=2b12b23,x2=b1b23\because x_1+2x_2=b_1 \implies \frac{b_1+2b_2}{3}+2x_2 = b_1 \implies 2x_2 = b_1 - \frac{b_1+2b_2}{3} = \frac{2b_1-2b_2}{3},x_2 = \frac{b_1-b_2}{3}

T1=write this for me!!!T^{-1} = \text{write this for me!!!}





T1MT=[1/32/31/31/3][1423][1211]T^{-1}MT = \begin{bmatrix}1/3&2/3\\1/3&-1/3\end{bmatrix}\begin{bmatrix}1&4\\2&3\end{bmatrix}\begin{bmatrix}1&2\\1&-1\end{bmatrix}

=[1/32/31/31/3][5251]=\begin{bmatrix}1/3&2/3\\1/3&-1/3\end{bmatrix}\begin{bmatrix}5&-2\\5&1\end{bmatrix} =[1532+23553213]=[5001]=\begin{bmatrix}\frac{15}{3}&\frac{-2+2}{3}\\\frac{5-5}{3}&\frac{-2-1}{3}\end{bmatrix}=\begin{bmatrix}5&0\\0&-1\end{bmatrix}

Quiz 1:

If we define [x1n+1x2n+1]=[1423][x1nx2n]\begin{bmatrix}x_{1_{n+1}}\\x_{2_{n+1}}\end{bmatrix}= \begin{bmatrix}1&4\\2&3\end{bmatrix}\begin{bmatrix}x_{1_n}\\x_{2_n}\end{bmatrix}

If we define [x10x20]=[35]\begin{bmatrix}x_{1_0}\\x_{2_0}\end{bmatrix} = \begin{bmatrix}3\\5\end{bmatrix}, what will x1nx_{1_n} be if nn\rightarrow \infty

(a) 0

(b) infinity

(c) 3

Quiz 2:

For n=10n = 10, x1=42317707x_1= 42317707, what will x2x_2 be?

(a) 84635418

(b) 0

(c) 42317709

Complex Numbers

i=1i = \sqrt{-1}

What will happen if eigenvalues are complex numbers? (Why I say "eigenvalues are" not "eigenvalue is?")

Quiz 3:

M = [1314]\begin{bmatrix}1&3\\-1&4\end{bmatrix}

eigenvalues are: 2.5000±0.8660i2.5000 \pm 0.8660i

eigenvectors are: [0.86600.4330±0.2500i]\begin{bmatrix}0.8660\\0.4330 \pm 0.2500i\end{bmatrix}

If we define [x1n+1x2n+1]=[1314][x1nx2n]\begin{bmatrix}x_{1_{n+1}}\\x_{2_{n+1}}\end{bmatrix}= \begin{bmatrix}1&3\\-1&4\end{bmatrix}\begin{bmatrix}x_{1_n}\\x_{2_n}\end{bmatrix}

What will happen if nn\rightarrow \infty

(a)spiral out

(b)spiral in

(c)stable limit cycle

Why?

(This will not be tested, but good to know)

Any complex number can be written as a+bi=C(cos(θ)+isin(θ))a+bi = C(cos(\theta)+isin(\theta))

C=a2+b2C = \sqrt{a^2+b^2} (Pythagorean theorem)

(C(cos(θ)+isin(θ)))2=C2(cos2(θ)+i2sin2(θ)+2icos(θ)sin(θ))=C2(cos2(θ)sin2(θ)+2icos(θ)sin(θ))(C(cos(\theta)+isin(\theta)))^2=C^2(cos^2(\theta)+i^2sin^2(\theta)+2icos(\theta)sin(\theta))=C^2(cos^2(\theta)-sin^2(\theta)+2icos(\theta)sin(\theta))

=C2(cos(2θ)+isin(2θ))=C^2(cos(2\theta)+isin(2\theta))

Some Example Questions

Question 2:

The Siberian tiger is an endangered subspecies of tiger that inhabits forests in Siberia and northern China. Set up a discrete-time model for a Siberian tiger population based on the assumptions below. If your model is linear, write down its matrix. If not, explain why not.

  • The population is divided into cubs (less than a year old), subadults (1–3 years old) and adults.
  • On average, adults have 1.5 cubs per year.
  • Bengal tiger cubs only remain cubs for one year. 52% of cubs die during this first year.
  • On average, 63% of subadults survive as subadults from one year to the next.
  • 20% of subadults mature into adults each year
  • 10% of adults die each year

We define three state variables

Cubs: CC, Subadults: SS, and Adult: AA

  • On average, adults have 1.5 cubs per year.

Cn+1=1.5AnC_{n+1} = 1.5A_{n}

  • Bengal tiger cubs only remain cubs for one year. 52% of cubs die during this first year.

All cubs are not cubs the next year, and (100%-52%)= 48% of them become subadults. Sn+1=0.48CnS_{n+1}= 0.48C_n

  • On average, 63% of subadults survive as subadults from one year to the next.

Sn+1=0.48Cn+0.63SnS_{n+1}=0.48C_n+0.63S_n

  • 20% of subadults mature into adults each year

An+1=0.2SnA_{n+1}= 0.2S_n

  • 10% of adults die each year

An+1=0.2Sn+0.9AnA_{n+1}= 0.2S_n+0.9A_n

[Cn+1Sn+1An+1]=[001.50.480.63000.20.9][CnSnAn]\begin{bmatrix}C_{n+1}\\S_{n+1}\\A_{n+1}\end{bmatrix} = \begin{bmatrix}0&0&1.5\\0.48&0.63&0\\0&0.2&0.9\end{bmatrix}\begin{bmatrix}C_n\\S_n\\A_n\end{bmatrix}

Question 2:

The following discrete-time matrix model describes the population of alligators in Florida, where the population has been divided into three life stages: juveniles (J), early adults (E), and adults (A).

[Jt+1Et+1At+1]=[0.60.450.680.210.4000.090.84][JtEtAt]\begin{bmatrix}J_{t+1}\\E_{t+1}\\A_{t+1}\end{bmatrix} = \begin{bmatrix}0.6&0.45&0.68\\0.21&0.4&0\\0&0.09&0.84\end{bmatrix}\begin{bmatrix}J_t\\E_t\\A_t\end{bmatrix}

  • The eigenvalues of the matrix in this model are 0.67, 0.96, and 0.21. Regardless of the initial state, what will happen to the alligator population in the long run? (Be as specific as possible.)
  • Find the dominant eigenvector of this model

[0.60.450.680.210.4000.090.84][0.960000.960000.96]\begin{bmatrix}0.6&0.45&0.68\\0.21&0.4&0\\0&0.09&0.84\end{bmatrix}-\begin{bmatrix}0.96&0&0\\0&0.96&0\\0&0&0.96\end{bmatrix}

= [0.36000.45000.68000.21000.5600000.09000.1200]\begin{bmatrix} -0.3600&0.4500&0.6800\\ 0.2100&-0.5600& 0\\ 0& 0.0900& -0.1200\end{bmatrix}

0.21x10.56x2=0    3x18x2=00.21x_1-0.56x_2 =0 \implies 3x_1-8x_2 = 0

0.09x20.12x3=0    3x24x3=00.09x_2-0.12x_3 = 0 \implies 3x_2 - 4x_3 =0

x1=32,x2=12,x3=9x_1 = 32, x_2 = 12, x_3 = 9 will solve.

eigenvector is [32129]\begin{bmatrix}32\\12\\9\end{bmatrix}

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