# Check whether the subsets $A, B$ of $M(2, \mathbb{R})$ are open or not and closed or not. [closed]

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.

1. 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 $$A_n$$ $$A_n= \left( {\begin{array}{cc} 0 & 1\\ -\frac{1}{n} & 0 \end{array} } \right)\to \left( {\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array} } \right)=A$$

1. 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.
• You mean that, $A$ is closed but not open and $B$ is open but not closed...?
– MAS
Commented Sep 20, 2015 at 19:51
• No, I mean $A$ is not closed (that it is not closed still does not mean it is closed ), and $B$ is closed. And because $A=B^c$, then it follows that $A$ is open (as a complement of a the closed set $B$) Commented Sep 20, 2015 at 20:48
• Thanks. @svetoslav Now, it is clear to me.
– MAS
Commented May 28, 2016 at 7:07

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.