Calculate the eigenvectors of this $2 \times 2$ matrix (problem because all equations result in $0=0$) 
What's the eigenvalues and eigenvectors of this matrix
  $B=\begin{pmatrix} 2 & 0\\  0 & 2 \end{pmatrix}$?

The characteristic polynomial is $\lambda^{2}-4\lambda+4=0$
The (double-)eigenvalue is $\lambda=2$
Now we want calculate the eigenvector with these, insert $\lambda=2$ here:
$$\begin{pmatrix}
2-\lambda & 0\\ 
0         & 2-\lambda
\end{pmatrix}= \begin{pmatrix}
0 & 0\\ 
0 & 0
\end{pmatrix}$$
So we have $\begin{pmatrix}
0 & 0\\ 
0 & 0
\end{pmatrix} \begin{pmatrix}
x\\ 
y
\end{pmatrix}=\begin{pmatrix}
0\\ 
0
\end{pmatrix}$
I: $0x+0y=0$
II: $0x+0y=0$
But what would be the eigenvector now?
Can I randomly choose one myself? As example, would this be correct?
Eigenvector is $v= \begin{pmatrix}
1\\ 
2
\end{pmatrix}$.

An additional question, how would you write the eigenspace?
 A: Yes, for this matrix every nonzero vector is an eigenvector with eigenvalue $2$.
It is a bit of a detour in this case to find that by solving the characteristic polynomial -- but it is certainly a valid way to proceed, until you have the experience to recognize immediately how a matrix of this form behaves.
Generally what you would like to find in this case is a basis for the eigenspace (which in this case is the entire $\mathbb R^2$), so you should choose two linearly independent vectors. The simplest and most boring choice would be $(^1_0)$ and $(^0_1)$, but you can certainly also choose $(^1_2)$ and, for example $(^7_3)$ (or anything else that is not parallel to $(^1_2)$).
(Since question actually asks "what are the eigenvectors", a more strictly correct answer would be "every nonzero vector is an eigenvector" but giving a basis for the eigenspace is conventional and may be the kind of answer that's expected anyway).
A: In fact, every non-zero vector $(x,y)$ is an eigenvector of $B$ to the eigenvalue $2$. This is because $Bx=2x$ holds for every vector $x$
A: Simply pick the eigenvectors $\binom{1}{0}$ and $\binom{0}{1}$. These satisfy the given equations canonically: they form a standard basis for the family of eigenvectors in fact.
Yes, you can randomly choose one yourself. Don't forget a second, linearly independent one too.
The eigenspace is the span of the eigenvectors.
