Why this transformation matrix $A$ has $\begin{pmatrix}0 \\ 1\end{pmatrix}$ as Eigenvector? I have the following transformation matrix:
$$
A=\begin{pmatrix}
1  & 0 \\
-1 & 4
\end{pmatrix}
$$
If I resolve to find the eigenvalues I get:
$$
\begin{vmatrix}
A-\lambda I
\end{vmatrix} = 0
$$
which leads to:
$$
\lambda_1 = 1;
\lambda_2 = 4
$$
Now if I try to calculate the eigenvectors
For $\lambda_1$ I get:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}0 \\ -x_1+3 x_2 \end{pmatrix}
$$
and for $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}
$$
I see (and can compute using a symbolic calculation program) that there are two eigenvectors:
$$
e1 = \begin{pmatrix}0 \\ 1\end{pmatrix}
$$
$$
e2 = \begin{pmatrix}3 \\ 1\end{pmatrix}
$$
I can easily see why $\begin{pmatrix}3 \\ 1\end{pmatrix}$ is an eigenvector for the eigenvalue $\lambda_2$.
But I am struggling to understand why $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is an eigenvector.
Could someone help me to understand why?

For $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \end{pmatrix}\Rightarrow x_1=0\Rightarrow v_2=\begin{pmatrix}0 \\ x_2 \end{pmatrix}=x_2 \begin{pmatrix}0 \\ 1 \end{pmatrix}
$$
So your second eigenvector is $\begin{pmatrix}0 \\ 1 \end{pmatrix}$
