yes/No :Is $K$ is dense in $\mathbb{M_2}({\mathbb{C}) } ?$

Question

Given $K=\{ A\in \mathbb{M}_2(\mathbb{C}) \mid A\text{ has no real eigenvalues} \}, $

is $K$ dense in $\mathbb{M_2}({\mathbb{C}) } ?$

My attempt : I know that set of all invertible matrix are dense so $K$ is dense in $\mathbb{M_2}({\mathbb{C}) }$.

Is it true?

Answer 1

The set $K = \{A\in M_n(\mathbb C): \ A \text{ has no real eigenvalues }\}$ is dense in $M_n(\mathbb C)$.

Take a matrix $B\in M_n(\mathbb C)$. Perform Schur decomposition: $B = QUQ^{-1}$ with unitary $Q$, $U$ upper triangular. $U$ and $B$ are similar, so the diagonal elements of $U$ are the eigenvalues of $B$ with multiplicity.

If $B$ and $U$ have real eigenvalues, then $U+tiI$ and $B+ tiI=Q(U+tiI)Q^{-1}$ have no real eigenvalues for all but finitely many $t\in \mathbb R$. For $t\to 0$, we have $B+itI \to B$. The matrix $B+itI$ has no real eigenvalues for almost all $t$, which proves the density result.

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The idea of the prove is to use the density of the 'unreal' numbers $\mathbb C\setminus \mathbb R$ in $\mathbb C$. The Schur decomposition is a convenient form to perturb the eigenvalues.