Closure of the set of all matrices with no real eigenvalues in $M_2 (\mathbb R)$ What is the closure of the set of all matrices with no real eigenvalues in $M_2 (\mathbb R)$.
 A: Hint: You might want to make use of the fact that the complex eigenvalues of a real matrix occur in conjugate pairs. For example, if $1-3i$ is an eigenvalue of $A \in M_2(\mathbb{R})$, then $1+3i$ is also an eigenvalue of $A$. 
A: Hint. The coefficients of the characteristic polynomial of a matrix can be written in terms of the entries of the matrix, and the condition that that characteristic polynomial have no real roots can be written as an inequality involving its coefficients. Make this explicit.
A: The characteristic polynomial of a $2\times2$ matrix $A$ is $$x^2-tr(A)x+\det(A)$$
That its eigenvalues are not real means that $tr(A)^2-4\det(A)<0$. Therefore, in the closure we can only expect to have matrices with $tr(A)^2-4\det(A)\leq0$. The matrices with $tr(A)^2<4\det(A)$ are already the ones with non-real eigenvalues. So, let us see which matrices satisfying $tr(A)^2-4\det(A)=0$ are in the closure.
If $tr(A)^2=4\det(A)$ then $$4x^2-4tr(A)x+4\det(A)=4x^2-4tr(A)x+tr(A)^2=(2x-tr(A))^2.$$ This means that the matrix has only one real eigenvalue. 
These are matrices of the form $$A=\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}ut&\pm u^2\\\mp t^2&-ut\end{pmatrix}$$
with $a,u,t$ real. Don't mind too much the form of the second matrix, what is important is that it has trace and determinant equal to zero. So, the terms in the diagonal add up to zero and their product is equal to the product of the elements in the antidiagonal.
Consider $$B_n=\begin{pmatrix}a&1/n\\-1/n&a\end{pmatrix}+\begin{pmatrix}ut&\pm u^2\\\mp t^2&-ut\end{pmatrix}$$
Then $tr(B_n)=2a$ and $\det(B_n)=a^2+u^2/n+t^2/n+1/n^2$. Therefore $tr(B_n)^2<4\det(B_n)$.
We have that $B_n\to A$.
Therefore all matrices with one eigenvalue are in the closure.
