Suppose characteristic polynomial of $\varphi : V \to V$, $V$ is over $\mathbb{R}$, is written as $\chi = (t - \lambda)^k(t - \bar\lambda)^k p(x)$, where $\lambda \in \mathbb{C} \backslash \mathbb{R}$ and $\lambda, \bar\lambda$ are exactly of multiplicity $k$. Show that $\dim \ker [(t- \lambda)(t - \bar\lambda)]\Bigg|_{t=\varphi} \leq 2k$.
To this point my only guess was to extend $V$ to vector space over complex numbers and show that $\dim \ker (\varphi - \lambda \mathrm{id})(\varphi - \bar\lambda \mathrm{id}) = \dim \ker (\varphi - \lambda \mathrm{id}) + \dim \ker (\varphi - \bar\lambda \mathrm{id}) \leq k + k = 2k$ but I couldn't prove nor disprove this fact.
I think the statement is true. Suppose $A\in\mathbb{R}^{n\times n}$ is a matrix that can represent $\varphi$. Then it suffices to show $n-rank\left(\left(A-\lambda I_n\right)\left(A-\bar{\lambda}I_n\right)\right)\le 2k$.
Now we apply the Jordan form. Suppose $J\in\mathbb{C}^{n\times n}$ is the Jordan form of $A$ with $PAP^{-1}=J$. Then $P\left(\left(A-\lambda I_n\right)\left(A-\bar{\lambda}I_n\right)\right)P^{-1}=\left(J-\lambda I_n\right)\left(J-\bar{\lambda}I_n\right)$. Let $B$ be the Jordan blocks of eigenvalue $\lambda$, $C$ the blocks of $\bar{\lambda}$ and $D$ the blocks of all other eigenvalue respectively. Then
$$\left(J-\lambda I_n\right)\left(J-\bar{\lambda}I_n\right)=\begin{pmatrix} \lambda I_n-B & & \\ & \lambda I_n-C & \\ & & \lambda I_n-D \end{pmatrix} \begin{pmatrix} \bar{\lambda}I_n-B & & \\ & \bar{\lambda}I_n-C & \\ & & \bar{\lambda}I_n-D \end{pmatrix} $$
$$=\begin{pmatrix} (\lambda I_n-B)(\bar{\lambda}I_n-B) & & \\ & (\lambda I_n-C)(\bar{\lambda}I_n-C) & \\ & & (\lambda I_n-D)(\bar{\lambda}I_n-D) \end{pmatrix} $$
Hence $rank\left(\left(J-\lambda I_n\right)\left(J-\bar{\lambda}I_n\right)\right)=rank\left((\lambda I_n-B)(\bar{\lambda}I_n-B)\right)+\left((\lambda I_n-C)(\bar{\lambda}I_n-C)\right)+\left((\lambda I_n-D)(\bar{\lambda}I_n-D)\right)\ge rank\left((\lambda I_n-D)(\bar{\lambda}I_n-D)\right)=n-2k$ (The last equality is because both $\lambda I_n-D$ and $\bar{\lambda}I_n-D$ is invertible, and hence their product is invertible.) This implies $n-rank\left(\left(A-\lambda I_n\right)\left(A-\bar{\lambda}I_n\right)\right)\le 2k$ as desired.