Vector differential equation, wrong eigenvector

Solve $$x' = \begin{bmatrix}-1 & 2 \\0 & -1 \end{bmatrix}x$$ Using the Eigenvalue method.

Where obviously $x$ is a vector equation.

Now the eigenvalue is $\lambda = -1$ repeated, so to find the first eigenvector I just solve

$\begin{bmatrix}0 & 2 \\0 & 0 \end{bmatrix}n = 0$, which conviently gives $n = \begin{bmatrix}1 \\ 0 \end{bmatrix}$,

Then we have to solve

$\begin{bmatrix}-1 & 2 \\0 & -1 \end{bmatrix}n = \begin{bmatrix}1 \\ 0 \end{bmatrix}$ for a new eigenvector.

I keep getting the system (where $n_1, n_2$ are components of this $n$ vector)

$\begin{cases}-n_1 + 2n_2 = 1\\ -n_2 = 0 \end{cases}$

Which gives $n = \begin{bmatrix}-1 \\0\end{bmatrix}$, which is wrong. My book says the second one should be $\begin{bmatrix}0 \\ 1/2\end{bmatrix}$ which makes no sense.

1 Answer

The second equation you need to solve is $(A-\lambda I)m = n$, which is

$\begin{bmatrix}0 & 2 \\0 & 0 \end{bmatrix}m = \begin{bmatrix}1 \\ 0 \end{bmatrix}.$

The first line gives $2m_2=1$, so $n=\begin{bmatrix}0 \\ 1/2\end{bmatrix}$ works.