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I was trying the following problem which was as follows:

Consider the first order system of linear equations:
$\frac{dX}{dt}=AX; \space A=\begin{pmatrix} 3 &2 \\ -2&-1 \end{pmatrix}; X=\space \begin{pmatrix} x_1(t)\\ x_2(t) \end{pmatrix}$.
Then which of the following options are correct?

  1. The coeffecient matrix $A$ has a repeated eigenvalue $\lambda =1.$
  2. There is only one linearly independent eigenvector $X_1=\space \begin{pmatrix} 1\\ -1 \end{pmatrix}$
  3. The general solution of the ODE is $(aX_1+bX_2)e^t,$ where $a,b$ are arbitrary constants and $X_1=\space \begin{pmatrix} 1\\ -1 \end{pmatrix},X_2=\space \begin{pmatrix} t\\ \frac{1}{2}-t \end{pmatrix}$

4.The vector in the option (3) given above are linearly independent.

My Attempt: Option (1) is true as the characteristic equation of $A$ is given by $(\lambda-1)^2=0$.
Option (2) also appears to be correct as $(A-I)X=0$ gives $X=c(1,-1)^t, c$ being a scalar.

Option (4) is also correct as $\begin{vmatrix} 1 & t\\ -1 & \frac{1}{2}-t \end{vmatrix} \neq 0$ and so the vectors $X_1,X_2$ are L.I.
But I am stuck on option (3) .

Can someone point me in the right direction? Thanks in advance for your time.

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I get different eigenvalues for $A$. – copper.hat Apr 19 '13 at 14:37
up vote 0 down vote accepted

In your notation you have $$ aX_1e^t+bX_2e^t. $$ To prove that this expression is the general solution you need to show two facts:

  • First, you need to show that both $X_1e^t$ and $X_2e^t$ solve the system
  • Second, you need to show that they are linearly independent.

It seems that you already answered second part, so you need just to plug these functions in and decide whether they are indeed solutions.

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