Check whether the subsets $A, B$ of $M(2, \mathbb{R})$ are open or not and closed or not. State whether each of the subsets $A$, $B$ of $M(2,\mathbb R)$ are open or not and closed or not :
(i) $A$ is the set of all matrices in $M(2,\mathbb R)$ such that neither eigenvalue is real.
(ii) $B$ is the set of all matrices in $M(2,\mathbb R)$ such that both eigenvalues are real.
 A: *

*Let $S_c$ be the set of all matrices $A$ in $M(2,\mathbb R)$ such that neither eigenvalue of $A$ is real.
Let $A_n\to A$ in some norm in $\mathbb R^{2\times 2}$ where $\left\{A_n\right\}\subset S_c$
$$A_n=
  \left( {\begin{array}{cc}
   a_n & c_n\\  
   d_n & b_n      
\end{array} } 
\right)
$$
$$A=
  \left( {\begin{array}{cc}
   a & c\\  
   d & b      
\end{array} } 
\right)
$$
Because all norms in finite dimensions are equivalent (in our case $\dim M(2,\mathbb R)=4$) we have also the convergence in $\mathbb R$ of $a_n\to a$, $b_n\to b$, $c_n\to c$, $d_n\to d$. Also $A_n$ has both its eigenvalues purely complex $\forall n\in\mathbb N$ means that its characteristic polynomial
$$\det(A_n-\lambda I)= (a_n-\lambda)(b_n-\lambda)-c_nd_n=\lambda^2-(a_n+b_n)\lambda +a_nb_n-c_nd_n=0$$
has $2$ complex roots: $D_n=(a_n-b_n)^2+4c_nd_n<0\,\,\forall n\in\mathbb N$. But when passing to a limit in $n\to\infty$ we get $(a-b)^2+4cd\leq 0$. The point is that in the limiting process it may happen that for some sequence $A_n$ the limiting matrix $A$ to poses $1$ (double) real eigenvalue. So the set $S_c$ is not closed.
Example: Consider the sequence of matrices $A_n$
$$
A_n=
  \left( {\begin{array}{cc}
   0 & 1\\  
   -\frac{1}{n} & 0      
\end{array} } 
\right)\to
  \left( {\begin{array}{cc}
   0 & 1\\  
   0 & 0      
\end{array} } 
\right)=A
$$


*Let $S_r$ be the set of all matrices $A$ in $M(2,\mathbb R)$ such that $A$ has its both eigenvalues real.  Let $A_n\to A$ in some norm in $\mathbb R^{2\times 2}$ where $\left\{ A_n\right\}\subset S_r$. Then for all $n\in\mathbb N$ the characteristic polynomial of $A_n$
$$\det(A_n-\lambda I)= (a_n-\lambda)(b_n-\lambda)-c_nd_n=\lambda^2-(a_n+b_n)\lambda +a_nb_n-c_nd_n=0$$
has its both roots in $\mathbb R$. This means $D_n=(a_n-b_n)^2+4c_nd_n\ge 0\,\,\forall n\in\mathbb N$. After letting $n\to\infty$ we see that $D=(a-b)^2+4cd\ge 0\Rightarrow$ the limiting matrix $A$ also has its eigenvalues in $\mathbb R$. So the set $S_r$ is closed.

A: Hint :
If $M = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$ and $I=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$, then 
$$\det(M-XI) = (a-X)(d-X)-bc = X^2-(a+d)X+ad-bc.$$ 
The eigenvalues of $M$ are the roots of the previous polynomial and there exists a classic rule to know when a polynomial of degree $2$ with real coefficients has real roots.
