Let $\begin{bmatrix} \dot x \\ \dot y \end{bmatrix}$$=$$\begin{bmatrix}-5 & -3 \\ 3 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$.
a) Find the general solution $\begin{bmatrix}x \\ y \end{bmatrix}$.
b)Find $λ$ and a basis for which the system reduces to ̇$\dot u=λu+v$ and $ ̇\dot v=λv$.
The eigenvalue $λ=-2$ gives us the eigenvector $\begin{bmatrix}-1 \\ 1 \end{bmatrix}$ and generalized eigenvector $\begin{bmatrix}1/3 \\ 0 \end{bmatrix}$.
For part a), the general solution $\begin{bmatrix}x \\ y \end{bmatrix}=c_1 e^{-2t}\begin{bmatrix}-1 \\ 1 \end{bmatrix}+c_2e^{-2t}\Bigg(t\begin{bmatrix}-1 \\ 1 \end{bmatrix}+\begin{bmatrix}1/3 \\ 0 \end{bmatrix}\Bigg)$
I'm not sure about part b) can anyone explain it?
$$\begin{bmatrix} \dot x \\ \dot y \end{bmatrix}=\begin{bmatrix}-5 & -3 \\ 3 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} \implies X'=AX$$ You have that $$ \begin{cases} u'=\lambda u + v \\ v'=\lambda v \end{cases} $$ $$ \implies \pmatrix { u \\ v}' =\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}\pmatrix {u \\ v } \implies U'=BU$$
B is just the Jordan matrix. Try to evaluate it now $$J=H^{-1}AH$$ Where H is the matrix with eigenvectors $$H=\pmatrix {-1 & 1/3 \\1 &0}$$