Prove $Av=\lambda v$ iff $\det[(A-\lambda_R I)^2 + \lambda_I^2 I ]=0$, for real $A$ Let $A$ be a real matrix. Given $\lambda\in\mathbb C$, we know $Av=\lambda v$ is equivalent to $\det(A-\lambda I)=0$.
If we convert this expression into one that only involves real numbers, as also discussed in this other question, we see that $Av=\lambda v$ with $\lambda=\lambda_1+i\lambda_2$ and $v=v_1+iv_2$ is equivalent to
$$\begin{cases}
  (A-\lambda_1 I)v_1 = - \lambda_2 v_2, \\
  (A-\lambda_1 I)v_2 = \phantom{-}\lambda_2 v_1,
\end{cases} \tag 1$$
Putting these two equations together, we see that they imply
$$\begin{cases}
  [(A-\lambda_1 I)^2 + \lambda_2^2 I ]v_1 = 0, \\
  [(A-\lambda_1 I)^2 + \lambda_2^2 I ]v_2 = 0,
\end{cases} \tag 2$$
which is then equivalent to $\dim\ker[(A-\lambda_1 I)^2 + \lambda_2^2 I ]>0$, and thus the condition
$$\det[(A-\lambda_1 I)^2 + \lambda_2^2 I ]=0.\tag 3$$
Does this hold in the other direction as well? In other words, assuming (3) holds for some $\lambda_1,\lambda_2\in\mathbb R$. Does this imply that there are $v_1,v_2$ not both equal to zero such that (1) holds?
 A: Let us write $P=X-\lambda\in \mathbb{C}[X]$. Then since 
$$P\bar{P}=(X-\lambda)(X-\bar{\lambda})=X^2-2\lambda_1X+(\lambda_1^2+\lambda_2^2),$$
your condition $(3)$ can be rephrased as: $P(A)\bar{P}(A)$ is singular; whereas condition $(1)$ is: $P(A)$ is singular.
Now since $A$ is a real matrix, $\bar{P}(A)=\bar{P}(\bar{A})=\overline{P(A)}$, so $\det(\bar{P}(A))=\overline{\det(P(A))}$, and finally 
$$\det(P(A)\bar{P}(A)) = \det(P(A))\det(\bar{P}(A)) = |\det(P(A))|^2.$$
So indeed $P(A)$ is singular if and only if $P(A)\bar{P}(A)$ is.
A: Noting
$$ (A-\lambda_1 I)^2 + \lambda_2^2 I=(A-(\lambda_1+i\lambda_2)I)(A-(\lambda_1-i\lambda_2)I)$$
one has
\begin{eqnarray}
\det ((A-\lambda_1 I)^2 + \lambda_2^2 I)&=&\det(A-(\lambda_1+i\lambda_2)I)\det(A-(\lambda_1-i\lambda_2)I)\\
&=&|\det(A-(\lambda_1+i\lambda_2)I)|^2.
\end{eqnarray}
So $\det ((A-\lambda_1 I)^2 + \lambda_2^2 I)=0$ implies $\det(A-(\lambda_1+i\lambda_2)I)=0$  and hence there is $v=v_1+iv_2\neq0$ such that
$$ (A-(\lambda_1+i\lambda_2)I)(v_1+iv_2)=0 $$
which is (1) after separating real and imaginary parts.
