Prove $Av=\lambda v$ iff $\det[(A-\lambda_R I)^2 + \lambda_I^2 I ]=0$, for real $A$

Question

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?

Answer 1

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.

Answer 2

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.