EigenVector, Eigenvalues, System of ODEs, and Solution Verifaction

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

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?

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

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: $$w=(w_1,w_2)=(w_1,-w_1)=w_1(1,-1)$$ $$w=(1,-1)$$
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.
• Is the vector $\pmatrix{3\\-1}$ a vector too? Or $\pmatrix{6\\-2}$ Apr 21, 2020 at 15:55