why this is correct: $\det(C+Di)$ is not zero, then there exists some real number $a$ such that $\det(C + a D)$ is not zero I wonder why the following statement is correct:
supposing $C$ and $D$ are two real matrix, if the determinant of the complex matrix $C + D i  $ is not zero, then there exists some real number $a$ such that $\det(C + a D)$ is not zero?
(where $i$ is the complex unit)
Thanks a lot.
 A: Consider the complex polynomial $p(t)=\det(C+tD)$. If it vanishes on $\mathbb{R}$, then it has infinitely many roots, so it is constant equal to $0$. In particular $p(i)=0$.
Note: there is no particular reason to assume $C,D$ to be real. They could be complex. But they need to be square, and of the same size.
Application: I suspect this is why you asked the question. If two real square matrices $A,B$ are similar in $M_n(\mathbb{C})$, then they are similar in $M_n(\mathbb{R})$. Indeed, the assumption gives us $C,D$ in $M_n(\mathbb{R})$ such that $P=C+iD$ is invertible in $M_n(\mathbb{C})$ and $PA=BP$. This yields $CA=BC$, $DA=BD$, and $\det(C+iD)\neq 0$. By the above argument, there exists $t\in\mathbb{R}$ such that $Q=C+tD$ is invertible. And then $QA=BQ$ is the desired similarity in $M_n(\mathbb{R})$.
A: Consider the polynomial $p(x):=\det(C+xD)$. It is a polynomial, as a big sum of many products with expressions like $c_{ij}+x\cdot d_{ij}$. (In particular, this will have real entries.)
We have that $p(i)\ne 0$. It follows that $p\ne 0$, so there is a real number (in fact, almost all) $a$, such that $p(a)\ne 0$.
