Jordan Blocks of Complex Congujate Eigenvalues of a Real Operator Denote the linear space of linear operators on the linear space $V$ with field $\mathbb{F}$ by $L(V)$ and the linear space of $n \times n$ matrices with entries in $\mathbb{R}$ by $\mathbb{R}^{n\times n}$. Let $T:V \to V$ be an operator on the linear space $V$ over $\mathbb{R}$. Let $C_T(x)=\det(x I - T)$ be its characteristic polynomial. The coefficients of $C_T$ are in $\mathbb{R}$ since by definition $C_T(x)=\det\left(\mathcal{M}_B^B(xI-T)\right)$ and all the entries of $\mathcal{M}_B^B(xI-T)$ are in $\mathbb{R}$, where $\mathcal{M}_B^B$ is a linear isomorphism between $L(V)$ and $\mathbb{R}^{n \times n}$ with $B$ being a basis for $V$. Consequently, if $\lambda \in \mathbb{C} - \mathbb{R}$ is a root of $C_T$ then $\bar\lambda$ is also a root of $C_T$ with the same algebraic multiplicity. Now, I want to show that
$$\dim \ker (\lambda I - T)^{(m)} = \dim \ker (\bar \lambda I - T)^{(m)}, \qquad m = 1,\dots,r \tag{1}$$
where $r$ is the algebraic multiplicity of $\lambda$ and $\bar \lambda$. According to the spectral decomposition theorem, we have $V = \cdots \oplus V_\lambda \oplus \cdots \oplus V_{\bar \lambda} \oplus \cdots$, where $V_{\lambda} = \ker (\lambda I - T)^{(r)}$ and $V_{\bar \lambda} = \ker (\bar \lambda I - T)^{(r)}$. Let $B=(\cdots, B_{\lambda},\cdots,B_{\bar \lambda},\cdots)$ be the corresponding basis of this decomposition. Equation $(1)$ literally means that the blocks $\mathcal{M}_{B_\lambda}^{B_{\lambda}}(T|_{V_{\lambda}})$ and $\mathcal{M}_{B_{\bar \lambda}}^{B_{\bar \lambda}}(T|_{V_{\bar \lambda}})$ have complex conjugate Jordan sub-blocks of the following form
\begin{align}
J_{\lambda} =
\begin{bmatrix}
\lambda &  1      & \cdots & 0       \\
0       & \lambda & \ddots & \vdots  \\
\vdots  & \vdots  & \ddots & 1       \\
0       & 0       & \cdots & \lambda
\end{bmatrix}_{d \times d}, \qquad
J_{\bar \lambda}=
\begin{bmatrix}
\bar \lambda &  1           & \cdots & 0            \\
0            & \bar \lambda & \ddots & \vdots       \\
\vdots       & \vdots       & \ddots & 1            \\
0            & 0            & \cdots & \bar \lambda
\end{bmatrix}_{d \times d}, \qquad
J_{\bar \lambda} = \overline{J_{\lambda}}
\end{align}
or more compactly,
$$\mathcal{M}_{B_{\bar \lambda}}^{B_{\bar \lambda}}(T|_{V_{\bar \lambda}}) = \overline{\mathcal{M}_{B_\lambda}^{B_{\lambda}}(T|_{V_{\lambda}})}$$
How can I prove equation $(1)$?
 A: The idea is that you can get the result by using two standard operations you are already familiar with: (i) transposition $^T$ (which you know doesn't change rank) and (ii) conjugate transposition $^*$.  These are both involutions.
Now consider arbitrary $B\in\mathbb C^{n\times n}$ where $\text{rank}\big(B\big)=r$ and use the fact

a matrix has rank $r$ iff it has some $r\times r$ submatrix with
nonzero determinant and for $m\gt r$ all $m\times m$ minors are zero

I.e. using two well chosen permutation matrices $PBP'=\begin{bmatrix} B_r  &* \\  * &* \end{bmatrix}$ has $B_r$, the leading $r\times r$ principal submatrix, invertible. (Note that $*$ denotes entries we don't care about; this nearly overloads notation for conjugate transpose but the context should make it clear.) So
$$\text{rank}\big(B\big) =\text{rank}\big(PBP'\big) =\text{rank}\big( B_r\big) =r $$
Since $B_r$ is invertible we know $B_r$ is similar (unitarily if you like) to an upper triangular matrix with no zeros on the diagonal.  Then $B_r^*$ is similar to a lower triangular matrix with no zeros on the diagonal so $B_r^*$ is invertible.
This tell us that $\big(PBP'\big)^*=(P')^*B^*P^*$ has a leading $r\times r$ principal submatrix with rank $r$.  It could conceivably have an even larger submatrix that has non-zero determinant so applying the blocked quote
$$\text{rank}\big(B\big)=r\leq \text{rank}\big((P')^*B^*P^*\big)=\text{rank}\big(B^*\big)\leq \text{rank}\big(B\big)$$
where the right hands side comes by re-running the argument on $B^*$ (i.e. making use of the involution).
Conclude: the conjugation map that sends $i\mapsto -i$ (aplied component-wise to a matrix) doesn't change rank of $B$, because it can be written as $\overline B=(B^*)^T$
Or if you prefer, we can conclude with:
\begin{align}  
\text{rank}\Big(\big(\lambda I -A\big)^m\Big)  
&=\text{rank}\Big(\big((\lambda I -A)^m\big)^*\Big)  \\
&=\text{rank}\Big(\big((\lambda I)^* -A^*\big)^m\Big) \\ 
&=\text{rank}\Big(\big(\overline \lambda I -A^T\big)^m\Big) \\  
&=\text{rank}\Big(\big((\overline \lambda I -A^T)^m\big)^T\Big) \\  
&=\text{rank}\Big(\big( (\overline\lambda I)^T -(A^T)^T\big)^m\Big) \\ 
&=\text{rank}\Big(\big(\overline \lambda I -A\big)^m\Big)
\end{align}
where, the third equality holds since $A$ has real components.
A: For any matrix $A$ with entries in $\mathbb{R}$ we have
$$(\bar \lambda I - A)^{(m)} = \Big((\overline{\lambda \bar I - \bar A)}\Big)^{(m)} =
\Big((\overline{\lambda I - A)}\Big)^{(m)} = \overline{(\lambda I - A)^{(m)}}.$$
Consequently, equation $(1)$ that we want to show is equivalent to
$$\dim \ker (\lambda I - A)^{(m)} = \dim \ker \overline{(\lambda I - A)^{(m)}}, \qquad m = 1,\dots,r$$
where $A:= M_B^B(T)$ is the matrix of $T$ with respect to some basis $B$ of $V$. It is enough to show that for any $m \times n$ matrix $P$ with entries in $\mathbb{C}$, the following holds.
$$\dim \text{col} P = \dim \text{col} \overline P \tag{$*$}$$
This means that complex conjugation does not change rank. This also implies that
$$\dim \ker P = \dim \ker \overline P,$$
if you apply the rank-nullity theorem to $P$ and $\overline P$.  A direct proof of $(*)$ is given here.
