Suppose that ${x}\choose{y}$ = $e^{-3t}$ ${-2}\choose{-1}$ is a solution to the system $\textbf {x' = Ax}$, where $\textbf{A}$ is a matrix with constant entries. Which of the following must be true?
a. -3 is an eigenvalue of A
b. ${4}\choose{2}$ is an eigenvector of A
c. The trajectory of this solution in the phase plane with axes x, y is part of a straight line.
d. $e^{-3t}$ ${-1}\choose{-2}$ is another solution to the same system
e. The phase portrait of the system is an attracting node
f. The system is stable
This is a star node, and hence $\lambda = -3$. (Hence $a$ is right.) Since the eigenvalue is negative, it is an attracting node (which makes $e$correct).
I graphed the parametric function, and the result was a straight line extending only in the 3rd quadrant. Hence $c$ is right.
I wasn't sure of a good mathematical way to get the matrix A, but by trial and error, got that A = [0, -6; -3, 3]. Checking for $Av = \lambda v$, I got the right answer, which makes me believe I have the right A. From this, trA = 3, and det A = -18. Checking against the rule of stability, we get that this system is not stable. (hence $f$ is not an answer). Again using $Av = \lambda v$ where $v$ = ${4}\choose{2}$, I conclude that b is also the right answer.
I'm not too sure about d, but taking the eigenvector [-1, -2] with eigenvector -3, and plugging it into the equation $(A-\lambda I) v$, which does not yield 0, I conclude d is wrong.
Hence my final choices are a, b, c, and e. Unfortunately, this turns out to be wrong—where might I have made mistakes?
There's no way you can determine the matrix $A$. All you know is one eigenvector and corresponding eigenvalue. You certainly do not know that you have an attractor, as the other eigenvalue might be positive. Nevertheless, a,b,c are correct.