Let $A$ be an $n \times n$ invertible matrix with complex entries and call $A = R + iJ$, where $R$ is the real part of $A$ and $J$ is the imaginary part of $A$. Show that there exist a $\lambda_0 \in \mathbb{R}$ s.t $R+ \lambda_0J$ is invertible. Futhermore, conclude that if $A$ and $B$ are matrices with real entries and they are similar in $\mathbb{C}$ then they are in $\mathbb{R}$.
I've tried to prove the first statement by contradiction, but what kind of things I can do if $\det(R+xJ)= 0$ for each $x \in \mathbb{R}$??
Also, I cannot see the conection with the fact that if $A$ and $B$ are similar in $\mathbb{C}$, then they are in $\mathbb{R}$.
Consider the complex polynomial $\lambda \mapsto \det(R + \lambda J)$. Note that it is non-zero when $\lambda = i$, so it is not the zero polynomial. Thus it has finitely many $0$s in the complex plane, which means we cannot have $\det(R + \lambda J) = 0$ for all $\lambda \in \Bbb{R}$.
Now, suppose that $A$ and $B$ are real matrices, and $P = R + iJ$ is a complex invertible matrix such that $$B = P^{-1}AP \iff PB = AP \iff RB + iJB = AR + iAJ.$$ Equating real and imaginary parts, \begin{align*} RB &= AR \\ JB &= AJ. \end{align*} By the previous exercise, we can find some $\lambda_0 \in \Bbb{R}$ such that $Q = R + \lambda_0 J$ is invertible. We therefore have, $$QB = RB + \lambda_0 JB = AR + \lambda_0 AJ = AQ \implies B = Q^{-1} AQ,$$ proving real-similarity.