Solving $x’=Ax$ for 2×2 matrix with repeated eigenvalues

Question

I am trying to find the general solution to the system $x' = Ax$. Here,

$$ A = \begin{pmatrix}
1 & -3\\
3 & -5
\end{pmatrix} $$

Finding the eigenvectors and eigenvalues, I found the eigenvalue of $-2$ to correspond to the eigenvector $ \begin{pmatrix}
1\\
1
\end{pmatrix} $

I am confused about how to proceed to finding the final solution here. Any guidance is greatly appreciated!

Answer 1

You have a deficient matrix and need to find a generalized eigenvector.

You have a double eigenvalue of $\lambda_{1,2} = -2$.

Solve for $v_2$ using the row-reduced-echelon-form (RREF) of $$[A-\lambda I]v_2 = [A + 2I]v_2 = v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

Spoiler

$$v_2 = \begin{pmatrix} \dfrac{1}{3} \\ 0 \end{pmatrix}$$