Question

Suppose 7 = AT, where A is the 2 x 2 matrix below. A= (1 1 1 3

(a) Determine the eigenvalues and eigenvectors of A.

(b) Express the general solution of t = Az in terms of real valued functions.

(c) Sketch the phase portrait of the system. Do not forget to label your axes.

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Answer #1

(a) We have the matrix A as

\small A=\begin{pmatrix} 1 & -1\\ 1 &3 \end{pmatrix}

The eigenvalues of A are simply the roots of its characteristic equation, which is

\small \det(A-xI)=\begin{vmatrix} 1-x & -1\\ 1 &3-x \end{vmatrix}=(1-x)(3-x)+1=x^2-4x+4=(x-2)^2

Hence the roots of the characteristic equation are \small x=2,2

Thus, the matrix has a single eigenvalue, 2, with multiplicity 2.

Now, we have to find the eigenvectors corresponding to this eigenvalue, which are given by

\small Av=2v\Rightarrow \begin{pmatrix} 1 & -1\\ 1 &3 \end{pmatrix}\begin{pmatrix} v_1\\ v_2 \end{pmatrix}=\begin{pmatrix} 2v_1\\ 2v_2 \end{pmatrix}\Rightarrow \begin{pmatrix} v_1-v_2 \\ v_1+3v_2 \end{pmatrix}=\begin{pmatrix} 2v_1\\ 2v_2 \end{pmatrix}

Now, equating the components together, we get the same equation :-

\small \Rightarrow \begin{matrix} v_1-v_2=2v_1\\ v_1+3v_2=2v_2 \end{matrix}\Rightarrow v_1+v_2=0 \Rightarrow v_1=-v_2

Thus, the general form of the eigenvector is given by

\small \begin{pmatrix} v_1\\ v_2 \end{pmatrix}=\begin{pmatrix} -v_2\\ v_2 \end{pmatrix}=v_1\begin{pmatrix} -1\\ 1 \end{pmatrix}

this completes part (a)

(b)

For repeated eigenvalues, the general solution of the system is given by

\small x=c_1e^{\lambda\:t}\eta+c_2e^{\lambda\:t}\left(t\eta+\rho \right)

where

\small \lambda =2

\small \eta = \begin{pmatrix} -1\\ 1 \end{pmatrix}

and \small \rho is a solution of the equation

\small \left(A-\lambda\:I\right)\rho=\eta

So,

\small \left(A-2I\right)\rho=\begin{pmatrix} -1\\ 1 \end{pmatrix}

Let rho = (p1,p2) so we have

\small \begin{pmatrix} -1 &-1 \\ 1& 1 \end{pmatrix}\begin{pmatrix} \rho_1\\ \rho_2 \end{pmatrix}=\begin{pmatrix} -1\\ 1 \end{pmatrix}\Rightarrow \begin{matrix} -\rho_1-\rho_2=-1\\ \rho_1+\rho_2=1 \end{matrix}\Rightarrow \rho_1+\rho_2=1

So, one possible solution is obtained by putting p2 = 0, so

\small \rho=\begin{pmatrix} 1\\ 0 \end{pmatrix}

Thus, we have the general solution of the system as

\small x=c_1e^{\lambda\:t}\eta+c_2e^{\lambda\:t}\left(t\eta+\rho \right)

\small \Rightarrow x=c_1e^{2t}\begin{pmatrix} -1\\ 1 \end{pmatrix}+c_2e^{2t}\left(t\begin{pmatrix} -1\\ 1 \end{pmatrix}+\begin{pmatrix} 1\\ 0\end{pmatrix} \right)

Simplifying we get

\small \Rightarrow x=c_1e^{2t}\begin{pmatrix} -1\\ 1 \end{pmatrix}+c_2e^{2t}\begin{pmatrix} 1-t\\ t \end{pmatrix}

which is our required solution.

(c) and the phase portrait of the system is

3.0 تی 20 1.0 0.0 ا -1.0 - 20 -3.0 - -3 - -1 1 اليا

where y is along the vertical axis and x is along the horizontal axis

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