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$\newcommand{\align}[1]{\begin{align} #1 \end{align}} \newcommand{\diffx}{\frac{dx}{dt}} \newcommand{\diffy}{\frac{dy}{dt}}\newcommand{\equation}[1]{\begin{equation} #1 \end{equation}}$

I am given the following system of differential equations:$\align{\diffx&=2x+3y-7\\ \diffy&=-x-2y+6}$ .

My first step was then to write the equation into the following matrix format. $\mathbf{X'}=\pmatrix{2&3\\-1&-2}\mathbf{X}+\pmatrix{7 \\ 6}$.

I then proceeded to identify my $\mathbf{A}$, from which I got the following: $\equation{\mathbf{A}=\pmatrix{2 &3\\ -1& -2}}$. From which I begin to do the following procedure, and this is where I begin to assume my error is. I do the indicated the step $\mathbf{A}-\lambda\mathbf{I}$

Here I get the following matrix $\equation{\pmatrix{2-\lambda &3 \\ -1 & -2-\lambda}}$. From there I calculate the determinant, and the following eigenvalues (or at least I think thats what they are called.

$\det(\mathbf{A}-\lambda\mathbf{I})=\lambda^2-1=>(\lambda-1)(\lambda+1)=0 \therefore \lambda=\pm1$

After arriving at these steps I substitute in the known values into the $\mathbf{A}-\lambda\mathbf{I}$ so for $\lambda=1$ i got the following matrix:<\p>

$$\equation{\pmatrix{1 & 3\\ -1 &-3}}$$

From here I set up my system of equations: $\align{k_1+3k_2&=0\\-k_1+-3k_2&=0}$

My Question

Is my work right thus far, and also how do I get my eigenvector from these values and do it for the other eigenvalue?

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  • $\begingroup$ Your eigenvalues are correct ....+1. For the eigenvector it's not a system you have two equations and two eigenvetors ..you calculate them separately. $\endgroup$ Apr 21, 2020 at 16:04

2 Answers 2

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For $\lambda =1$ you have the matrix: $$\pmatrix{1 & 3\\ -1 &-3}\pmatrix {v_1 \\v_2}=\pmatrix {0\\0}$$ You find the eigenvector for the eigenvalue $\lambda =1$ $$v_1+3v_2=0 \implies v_1=-3v_2$$ You can choose for example: $$ v=(v_1,v_2)=(-3v_2,v_2)=v_2(-3,1)$$ $$v=(-3,1)$$ Do the same for the other eigenvalue. For $\lambda =-1$ you have the matrix: $$\pmatrix{ 3& 3\\ -1 &-1}\pmatrix {w_1 \\w_2}=\pmatrix {0\\0}$$ $$w_1+w_2=0 \implies w_1=-w_2$$ You can choose for example the eigenvector: $$ w=(w_1,w_2)=(w_1,-w_1)=w_1(1,-1)$$ $$ w=(1,-1)$$

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  • $\begingroup$ Yay i solved my first problem of this type, and got it right thanks +1 $\endgroup$ Apr 21, 2020 at 16:50
  • $\begingroup$ Thta's great good job.. @EnlightenedFunky $\endgroup$ Apr 21, 2020 at 16:52
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If $k_2=1$, then $k_1=-3$, hence $\begin{bmatrix} -3 \\ 1\end{bmatrix}$ is an eigenvector.

Similarly, if $\lambda_2 = -1$, then we have $3k_1+3k_2=0$, again, if $k_2=1$ then $k_1=-1$, hence $\begin{bmatrix} -1 \\ 1\end{bmatrix}$ is another eigenvector.

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  • $\begingroup$ Is the vector $\pmatrix{3\\-1}$ a vector too? Or $\pmatrix{6\\-2}$ $\endgroup$ Apr 21, 2020 at 15:55
  • $\begingroup$ any non-zero multiples of an eigenvector is another eigenvector. $\endgroup$ Apr 21, 2020 at 16:00

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