I am stuck on a problem and I do not know how to obtain the eigenvectors:
$\frac{dY}{dt}=\bigl(\begin{smallmatrix} -2&0\\ -3&1 \end{smallmatrix} \bigr)Y$
Work: I obtained the eigenvalues $-2,1$ and therefore the origin must be a saddle. However, when I try and obtain the eigenvector, I get the system:
$-4x=0$
$-3x-y=0$.
I am unable to obtain anything from this.
You set up and solve, using RREF:
$$[A - \lambda_i I]v_i = 0$$
The eigenvalues are: $\lambda_1 = -2, \lambda_2 = 1$
The corresponding eigenvectors are:
$$(v_1 | v_2) = \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array} \right)$$
For $\lambda_1 = -2$, we have the RREF of $[A +2I]v_1 = 0$ as:
$$ \left( \begin{array}{cc} 1 & -1 \\ 0 & 0 \\ \end{array} \right)v_1 = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$
This means we have $a = b$, so we can choose $a = b = 1$.
For $\lambda_2 = 1$, we have the RREF of $[A -I]v_2 = 0$ as:
$$ \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right)v_2 = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$
This means we have $a = 0 $, so we can choose $b = 1$.
Notice how, in both cases, that eigenvector chosen solves the matrix above? Also note that you cannot have a zero eigenvector.