\[\newcommand{\arr}[1]{\underline{\underline{#1}}}\]
\[\newcommand{\vec}[1]{\underline{#1}}\]
\[\require{mhchem}\]

Eigenvalues/Eigenvectors continued, and application of eigenvalues to difference equations#

\(\underline{Ex}\): Symmetric matrix (\(\lambda\) always real)#

  • A real square matrix \(\arr{A}=[a_{jk}]\) is symmetric if \(\arr{A}^T = \arr{A}\implies\) thus \(a_{kj}=a_{jk}\)
    \(\arr{A}=\begin{bmatrix} -5&2\\2&-2 \end{bmatrix}\)
    Eigenvalues:

(53)#\[\begin{align} |\arr{A}-\lambda\arr{I}|=0\\ \left|\begin{array}{} -5-\lambda&2 \\2&-2-\lambda \end{array} \right|=0\\ &=(-5-\lambda)(-2-\lambda)-4\\ &=10+5\lambda+2\lambda+\lambda^2-4\\ &=\lambda^2+7\lambda+6\\ &=(\lambda+6)(\lambda+1)\\ \implies \lambda_1=-6 ;\hspace{1cm} \lambda_2 = -1 \end{align}\]

  • The eigenvalues of symmetric matrices are always real.
    Eigenvectors:
    For \(\lambda_1=-6\),

(54)#\[\begin{align} (\arr{A} + 6\arr{I})\vec{x}^{(1)} = \vec{0}\\ \begin{bmatrix} 1&2\\2&4 \end{bmatrix} \rightarrow \left[ \begin{array}{rr|r} 1&2&0\\0&0&0\end{array} \right]\\ \implies x_1 + 2x_2 = 0\\ x_1 = -2x_2\\ x^{(1)}=[-2, 1]^T \end{align}\]

For \(\lambda_2=-1\),

(55)#\[\begin{align} (\arr{A} + \arr{I})\vec{x}^{(2)} = \vec{0}\\ \begin{bmatrix} -4&2\\2&-1 \end{bmatrix} \rightarrow \left[ \begin{array}{rr|r} 0&0&0\\2&-1&0\end{array} \right]\\ \implies 2x_1 - x_2 = 0\\ x_2 = 2x_1\\ x^{(1)}=[1, 2]^T \end{align}\]

Eigenspace:

(56)#\[\begin{align} \vec{x}=\vec{0}; && \lambda=-6, \vec{x} = \begin{bmatrix} -2\\1 \end{bmatrix}; && \lambda=-1, \vec{x} = \begin{bmatrix} 1\\2 \end{bmatrix} \end{align}\]

\(\underline{Ex}:\) A skew symmetric matrix (\(\lambda=0\) or complex)#

\(\rightarrow \arr{A}^T = -\arr{A}\) \(\hspace{0.5cm}(a_{ij}=-a_{ji})\)
\(\arr{A} = \begin{bmatrix} 0&9&-12 \\ -9&0&20 \\ 12&-20&0 \end{bmatrix}\)

  • Find eigenvalues by solving \(|\arr{A} - \lambda \arr{I}| = 0\)

(57)#\[\begin{align} \left|\begin{array}{} -\lambda&9&-12 \\ -9&-\lambda&20 \\ 12&-20&-\lambda \end{array} \right| = 0\\ -\lambda \left|\begin{array}{} -\lambda&20\\-20&-\lambda \end{array}\right| -9 \left|\begin{array}{} -9&20\\12&-\lambda \end{array}\right| -12 \left|\begin{array}{} -9&-\lambda\\12&-20 \end{array}\right| = 0\\ -\lambda(\lambda^2+400) -9 (9\lambda - 240) -12 (180 + 12\lambda) = 0\\ -\lambda^3 - 400\lambda - 81\lambda + 2160 - 2160 - 144\lambda = 0\\ -\lambda^3 - 625\lambda = 0\\ \lambda(\lambda^2 + 625) = 0\\ \end{align}\]
(58)#\[\begin{align} \lambda_1= 0 && \lambda_2 = 25i && \lambda_2 = -25i \end{align}\]

  • The eigenvalues of skew-symmetric matrices are always complex or zero.

  • Find eigenvectors:

    1. Find \(\vec{x}^{(1)}\) from \((\arr{A} - 0\arr{I}) = \vec{0}\)

    (59)#\[\begin{align} \left[ \begin{array}{rrr|r} 0&9&-12&0 \\ -9&0&20&0 \\ 12&-20&0&0 \end{array} \right] \end{align}\]

    Swap rows:

    (60)#\[\begin{bmatrix} 12&-20&0 \\ -9&0&20 \\ 0&9&-12 \end{bmatrix}\]

    \(R_3 = \frac{1}{3}R_3\) ; \(R_1 = \frac{1}{4}R_1\):

    (61)#\[\begin{bmatrix} 3&-5&0 \\ -9&0&20 \\ 0&3&-4 \end{bmatrix}\]

    \(R_2 = R_2 +3R_1\), \(R_3 = 5R_3\):

    (62)#\[\begin{bmatrix} 3&-5&0 \\ 0&-15&20 \\ 0&15&-20 \end{bmatrix}\]

    \(R_3 = R_3 + R_2\):

    (63)#\[\begin{bmatrix} 3&-5&0 \\ 0&-15&20 \\ 0&0&0 \end{bmatrix}\]

    $R_2 = \frac{1}{5} R_2$\

    (64)#\[\begin{align} \left[ \begin{array}{rrr|r} 3&-5&0&0 \\ 0&-3&4&0 \\ 0&0&0&0 \end{array} \right] \end{align}\]
    (65)#\[\begin{align} \implies 3x_1 - 5x_2 = 0\\ -3x_2 + 4x_3 = 0\\ \implies x_2 = \frac{4}{3}x_3\\ x_1 = \frac{20}{9}x_3\\ \end{align}\]
    (66)#\[\begin{align} \lambda_1 = 0, && \arr{x}^{(1)} = \begin{bmatrix}20\\12\\9 \end{bmatrix} \end{align}\]

    Note these are the off-diagonal terms
    \(\vec{x}^{(2)}\) and \(\vec{x}^{(3)}\) will be complex. Practice finding them @ home.

  • Eigenspace of \(\arr{A}\) is:

(67)#\[\begin{align} \vec{x} = \vec{0}; && 0, \begin{bmatrix} 20 \\ 12 \\9 \end{bmatrix}; && 25i, \begin{bmatrix} -0.33 + 0.55i \\ -0.20 - 0.92i \\ 1 \end{bmatrix} && -25i, \begin{bmatrix} -0.33 - 0.55i \\ -0.20 + 0.92i \\ 1\end{bmatrix} \end{align}\]

Note the complex conjugate vectors.

Final Example with Triangular Matrix:#

\(\arr{A} = \begin{bmatrix} 1&0&0 \\ -9&2&0 \\ 12&1&-3 \end{bmatrix} \rightarrow\) lower triangular matrix, \(a_{ij} = 0\) if \(j>1\)

  • Eigenvalues from \(|\arr{A}-\lambda \arr{I}| = 0\)

(68)#\[\begin{align} \left| \begin{array}{} 1-\lambda&0&0 \\ -9&2-\lambda&0 \\ 12&1& -3-\lambda \end{array} \right| = 0\\ (1-\lambda)\left| \begin{array}{} 2-\lambda&0\\1&3-\lambda \end{array}\right|=0\\ (1-\lambda)(2-\lambda)(3-\lambda)=0\\ \end{align}\]
(69)#\[\begin{align} \implies \lambda_1 = 1,&& \lambda_2 = 2,&& \lambda_3 = -3 \end{align}\]

  • For upper or lower triangular matrices, eigenvalues will be diagonal elements.

One Application of the Eigenvalue Problem#

  • “Difference” Equations (Recurrance Relations)

    • Can be used to solve problems in population dynamics, digital signal processing and economics.

    • We will use one here to determine how quickly people die from a plague.

  • First we must understand that the eigenvectors of any n-dimentional problem form a basis for \(R^n\). In 3D space for example, we often think of \([1, 0, 0], [0, 1, 0]\) & \([0, 0, 1]\) as basis vectors of \(R^n\). Eigenvectors also form a basis because they are always linearly independent.

  • Because eigenvectors form a basis, we can write any vector, \(\vec{z}\), as

(70)#\[\begin{align} \vec{z} = c_1 \vec{x_1}+ c_2 \vec{x_2} + ... c_n \vec{x_n} && \text{for scalars $c_1 ... c_n$}\\ \end{align}\]
(71)#\[\begin{align} e.g. [2,4,3] &= 2[1, 0, 0] + 4[0, 1, 0] + 3[0, 0, 1]\\ &=-25[-2, -1, -1] -13[3, 3, 1] -9[1, -2, 1] \end{align}\]
(72)#\[\begin{align} \therefore \arr{A}\vec{z} &= \arr{A}(c_1 \vec{x_1} + c_2 \vec{x_2}+...c_n \vec{x_n})\\ &= c_1\arr{A}\vec{x_1} + c_2\arr{A}\vec{x_2} + ... c_n\arr{A}\vec{x_n} \end{align}\]

But, since \(\arr{A}\vec{x_n} = \lambda_n\vec{x_n}\), we can greatly simplify:

(73)#\[\begin{align} \arr{A}\vec{z} = c_1\lambda_1\vec{x_1} + c_2\lambda_2\vec{x_2} + ... + c_n\lambda_n \vec{x_n} \end{align}\]

\(\therefore\) If we know \(\arr{A}\), finding its eigenvalues and vectors will help us find \(\vec{z}\)

  • This becomes interesting when wanting to multiply an eigenvector \(\vec{x}\) by powers of \(\arr{A}\). For example:

(74)#\[\begin{align} \arr{A}^2\vec{x} = \lambda^2\vec{x} && (\arr{A}\arr{A}\vec{x} = \arr{A}\lambda\vec{x} = \lambda\arr{A}\vec{x} = \lambda\lambda\vec{x} = \lambda^2\vec{x})\\ \arr{A}^3\vec{x} = \lambda^3\vec{x}\\ \arr{A}^k\vec{x} = \lambda^k\vec{x} \end{align}\]

then, \(\arr{A}^k\vec{z} = c_1\lambda_1^k\arr{x_1} + c_2\lambda_2^k\vec{x_2} + ... + c_n\lambda_n^k\vec{x_n}\)

  • We are interested in equations of the form \(\arr{A}^k\vec{z}\) because it can tell us how some initial state represented by vector \(\vec{z}\) changes on a recurrant basis. For example, if \(\arr{z_0}\) represents the initial state of a population and \(\arr{A}\) tells us how the population changes each year then after 1 year:

(75)#\[\begin{align} \vec{z_1} = \arr{A}\vec{z_0} \end{align}\]

And after 2 years:

(76)#\[\begin{align} \vec{z_2} &= \arr{A}\vec{z_1} = \arr{A}\arr{A}\vec{z_0}\\ \vec{z_2} &= \arr{A}^2\vec{z_0} \end{align}\]

And after k years:

(77)#\[\begin{align} \vec{z_k} = \arr{A}^k\vec{z_0} &&\rightarrow \text{this type of equation is called a difference equation} \end{align}\]

\(\textbf{Problem:}\) A fearsome new strain of Ebola strikes the island of Niihau, HI with population of 200. People can be categorized as healthy, sick or dead. Each year, the plague causes 60% of healthy people to get sick and another 10% of healthy people to die. Only 30% stay healthy. This strain of Ebola is difficult to cure, and so 60% of sick people die each year, 20% become healthy and 20% will remain sick.

  • Determine the equation that predicts how many people are healthy, sick and dead after k years.

  • How many years will it take before only 20 people remain alive on Niihau?

  • Begin by setting up variables and equations:
    Let \(z_1(k)\) = number of healthy people after k years
    \(z_2(k)\) = number of sick people after k years
    \(z_3(k)\) = number of dead people after k years
    \(\implies\) 3 types of people, 3 dimensions

  • Then info in problem tells us:
    HEALTHY

(78)#\[\begin{align} z_1(k+1) = 0.3z_1(k) + 0.2 z_2(k) \end{align}\]

  • In a given year \((k+1)\), the number of healthy people equals 30% of the healthy people from the previous year \((k)\) + 20% of the sick people from the previous year \((k)\).

SICK

(79)#\[\begin{align} z_2(k+1) = 0.6 z_1(k) + 0.2 z_2(k) \end{align}\]

DEAD

(80)#\[\begin{align} z_3(k+1) = 0.1 z_1(k) + 0.6 z_2(k) + z_3(k)\\ & \rightarrow \text{All dead people stay dead} \end{align}\]

So if \(\vec{z_k} = \begin{bmatrix} z_1(k) \\ z_2(k) \\ z_3(k) \end{bmatrix}\), then this problem is asking us to solve the difference equaiton

(81)#\[\begin{align} \vec{z_k} = \arr{A}\vec{z_{k-1}} = \arr{A}^k\vec{z_0} \end{align}\]

  • Let’s write the initial population vector in terms of eigenvalues and vectors of \(\arr{A}\).

(82)#\[\begin{align} \vec{z_k} = c_1\lambda_1^k\vec{x_1} + c_2\lambda_2^k\vec{x_2} + c_3\lambda_3^k\vec{x_3} \end{align}\]

  • Need to find eigenvalues and eigenvectors of \(\arr{A} = \begin{bmatrix} 0.3&0.2&0 \\ 0.6&0.2&0 \\ 0.1&0.6&1 \end{bmatrix}\)

(83)#\[\begin{align} \lambda_1 = -0.1, && \vec{x}^T = [1, -2, 1]\\ \lambda_2 = 0.6, && \vec{x}^T = [-2, -3, 5]\\ \lambda_1 = 1, && \vec{x}^T = [0, 0, 1] \end{align}\]

  • Finally we can find the \(c_n\) values by expressing the initial population vector, \(\vec{z_0} = \begin{bmatrix} 200\\0\\0 \end{bmatrix}\) in terms of the eigenvector basis.

(84)#\[\begin{align} \vec{z_0} &= \frac{600}{7}\vec{x_1} - \frac{400}{7}\vec{x_2} + 200 \vec{x_3}\\ \begin{bmatrix}200\\0\\0\end{bmatrix} &= \frac{600}{7}\begin{bmatrix}1\\-2\\1\end{bmatrix} - \frac{400}{7}\begin{bmatrix}-2\\-3\\5\end{bmatrix} + 200 \begin{bmatrix}0\\0\\1\end{bmatrix} \end{align}\]

Here the coefficients represent \(c_1, c_2\) and \(c_3\) respectively.

(85)#\[\begin{align} \therefore \vec{z_k} = \frac{600}{7}(-0.1)^k\begin{bmatrix}1\\-2\\1\end{bmatrix} - \frac{400}{7}(0.6)^k\begin{bmatrix}-2\\-3\\5\end{bmatrix} + 200(1)^k\begin{bmatrix}0\\0\\1\end{bmatrix} \end{align}\]

  • Finally, to determine how long it will take to have only 20 people remaining, we can plug in some values of \(k\).

  • Rounded to the nearest person:

(86)#\[\begin{align} \vec{z_0} = \begin{bmatrix} 200\\0\\0 \end{bmatrix} && \vec{z_1} = \begin{bmatrix} 60\\120\\20 \end{bmatrix}\\ \vec{z_2} = \begin{bmatrix} 42\\60\\98 \end{bmatrix} && \vec{z_3} = \begin{bmatrix} 25\\37\\138 \end{bmatrix}\\ \vec{z_4} = \begin{bmatrix} 15\\22\\163 \end{bmatrix} && \vec{z_5} = \begin{bmatrix} 9\\13\\178 \end{bmatrix}\\ \vec{z_6} = \begin{bmatrix} 5\\8\\187 \end{bmatrix} \end{align}\]

  • After 6 years, only 13 people remain (less than 10% of original population)