The set of all matrices in $M_2(r)$ such that both of their Eigenvalues are purely imaginary is closed or not? The set of all matrices in $M_2(r)$ such that both of their Eigenvalues are purely imaginary is not closed.
Can I have an sequence which will converge to a matrix which does not belong to this set. If it is not closed it has to have because it is Hausdorff.
 A: Take 
$$
M(n)=\begin{bmatrix}0&\frac{1}{n}\\-\frac{1}{n} &0\end{bmatrix}
$$
with eigenvalues 
$$
\pm \frac{i}{n}
$$
but limit the zero matrix.
Edit: I interpreted $0$ not to be purely imaginary. If instead you do consider it to be imaginary, then the set of $2\times 2$ matrices with purely imaginary eigenvalues is closed by Daniel Schepler's point, since the space is given by 
$$
\left\{\text{Tr}^{-1}(\{ 0\})\right\}\cap \left\{\text{det}^{-1}((-\infty,0])\right\}
$$
A: The characteristic polynomial for
$$
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
$$
is
$$
X^2-(a+d)X+(ad-bc)
$$
In order for the eigenvalues to have zero real part, we need $a+d=0$. The condition for no real roots is then $ad-bc>0$.
Now it depends on whether you consider $0$ to be purely imaginary or not. If not, the set is not closed. Otherwise it is, because it is the set of matrices of the form
$$
\begin{bmatrix} a & b \\ c & -a \end{bmatrix}
$$
with $a^2+bc\le0$.
Under most conventions, $0$ is considered to be purely imaginary, just like the zero matrix is considered both symmetric and antisymmetric.
