Finding the eigenvalues and eigenvectors with each eigenvalue, solving the general solution with initial conditions.

Consider the system $x'_1 = x_1 + 2x_2$ and $x'_2 = 3x_1 + 2x_2$

If we write in matrix from as $X' = AX$, then

a) $X =$

b) $X' =$

c) $A =$

d) Find the eigenvalues of A.

e) Find eigenvectors associated with each eigenvalue. Indicate which eigenvector goes with which eigenvalue.

f) Write the general solution to the system.

g) Find the specific solution that satisfies the initial conditions $x_1(0) = 0$ and $x_2(0) = -4$

Ok so here are my solutions so far

a) $X = \vec{X} = (^{x_1}_{x_2})$

b) $X' =$ \begin{bmatrix} (1-\lambda) & 2 \\ 3 & (2-\lambda) \end{bmatrix}

c) $A =$ \begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix}

d) $\lambda_1 = -1$ and $\lambda_2 = 4$

Items $\textbf{(a)}, \textbf{(b)}$ and $\textbf{(c)}$ are just meant to see if you can write the system in matrix form. That is, if you can recognize that: $$\begin{cases} x_1' = x_1 + 2x_2 \\ x_2' = 3x_1 + 2x_2 \end{cases} \iff \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

That being said, your item $\textbf{(b)}$ is wrong: you're way ahead of yourself there. That matrix is $A - \lambda\,{\rm Id}_2$, not $X'$.

For item $\textbf{(d)}$, I'll trust you that $\lambda_1 = -1$ and $\lambda_2 = 4$. For $2\times 2$ matrices, the characteristic polynomial is $p(\lambda) = \lambda^2 -{\rm tr}(A)\,\lambda +\det(A)$, in general (prove this as an extra exercise!).

For item $\textbf{(e)}$, remember that ${\bf v}_1$ is an eigenvector associated to $\lambda_1$ if and only if $A{\bf v}_1 = \lambda_1{\bf v}_1$, that is, if and only if $(A-\lambda_1\,{\rm Id}_2){\bf v}_1 = 0$. This means that ${\bf v}_1 \in \ker(A - \lambda_1\,{\rm Id}_2)$. Same goes for $\lambda_2$.

So:

• for $\lambda_1 =-1$ we solve $(A+{\rm Id}_2){\bf v} =0$ in general and pick a nice ${\bf v}_1$ as a particular solution. In details: $$\begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies a = -b$$ If everything is ok, then $\det(A+{\rm Id}_2) = 0$, and this is indeed the case. You can always compute the determinant to double-check your work so far. So $${\bf v} = \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} a \\ -a \end{bmatrix} = a \begin{bmatrix} 1 \\ -1 \end{bmatrix},$$ so that we can pick $${\bf v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.$$ Actually you can pick any multiple if the above vector to be our ${\bf v}_1$, it will work. The first column of the fundamental matrix is $$e^{\lambda_1t}{\bf v}_1 = e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix}.$$

• for $\lambda_2 = 4$ you will do the same as above, and you should get: $${\bf v}_2 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}.$$ Doing a straightforward computation, your candidate for ${\bf v}_2$ must contain some fractions. Remember now what I've said before: as long as my ${\bf v}_2$ and your ${\bf v}_2$ are multiples, everything is going to be fine. Pay attention to that. So the second column of the fundamental matrix is: $$e^{\lambda_2t}{\bf v}_2 = e^{4t}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2e^{4t} \\ 3e^{4t} \end{bmatrix}.$$

So our fundamental matrix is $$F(t) = \begin{bmatrix} e^{-t} & 2e^{4t} \\ -e^{-t} & 3e^{4t} \end{bmatrix}.$$

This means that the general solution is given by $$X(t) = F(t) C \iff \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = \begin{bmatrix} e^{-t} & 2e^{4t} \\ -e^{-t} & 3e^{4t} \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \iff \begin{cases} x_1(t) = c_1 e^{-t} + 2c_2e^{4t} \\ x_2(t) = -c_1e^{-t} + 3c_2e^{4t} \end{cases},$$ where $c_1,c_2 \in \Bbb R$. If you manage to understand everything so far, solving the final IVP will pose no problem: just plug in the values of $t$ and solve for $c_1$ and $c_2$. Now it's on you!

For (b), we have $X' =\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}'=\begin{bmatrix}x_1' \\ x_2' \end{bmatrix}$.

Edit:

To determine eigenvalues, solve the characteristic equation $f(\lambda)=\det(A- \lambda I)=0$.

To find the corresponding eigenvector for each eigenvalue $\lambda$, solve the system $(A-\lambda I)X=0$. After finding the eigenvalues $\lambda_1$ and $\lambda_2$, with corresponding eigenvectors $\xi_1$ and $\xi_2$, your solution will be

$$X(t)=c_1\xi_1 e^{\lambda_1 t} + c_2 \xi_2 e^{\lambda_2 t}.$$

You can use the given inital condition $X(0) =\begin{bmatrix}1\\-2\end{bmatrix}$ to solve for $c_1$ and $c_2$.

• Are you saying my part $b)$ is incorrect? Also, I already have the eigenvalues, I don't need help with that. What I need help with is how to solve the system for $d)$..How do I apply what you're saying to do. – Yusha Apr 21 '15 at 15:02
• Yes, it is incorrect. I'm not sure why you downvoted my answer ... – MathMajor Apr 23 '15 at 0:16
• Edit your answer and just put "Edit" or something so I can upvote it back up. – Yusha Apr 26 '15 at 1:59
• Because It won't let me upvote the answer unless you edit it. It's locked or something. – Yusha Apr 26 '15 at 1:59
• @user3325915 Okay, thank you. – MathMajor Apr 26 '15 at 2:00