Prove that the size and the number of Jordan block's of $\lambda$ and $\bar{\lambda}$ are the same. Prove that the size and the number of Jordan block's of $\lambda$ and $\bar{\lambda}$ are the same where $T$ is a real operator on $V$ a finite dimensional space.
I know that the main is to show that $\dim \ker (T - \lambda I)^j = \dim \ker (T - \bar{\lambda}I)^j$, for every $j$.
But, how to do this?
 A: I am assuming that this is for a real linear transformation over a finite dimensional, vector space (If the transformation were complex, it would not be true at all). 
Let $\bar{v}$ denote term-by-term conjugation of the vector $v$, and let $\bar{T}$ denote term by term conjugation of the matrix representation of $T$. If you don't know this already, it is pretty easy to show using the properties of the complex conjugate that: $\overline{T(v)} = \overline{T}\bar{v}$ and  that for two linear transformations, $T$, $S$:
$\overline{T}\overline{S} = \overline{TS}$ 
$\overline{T} + \overline{S} = \overline{T+S}$
We know that if $v \in \mathrm{Null}(T-\lambda I)^j$, then $(T-\lambda I)^j(v) = 0$. 
Then we apply conjugation:
$\overline{(T-\lambda I)^j(v)} = 0$
$(\overline{T} -\overline{\lambda I})^j(\overline{v}) = 0$.
We must have that $T$ has all real entries by assumption:
$(T - \overline{\lambda}I)^j(\overline{v}) = 0$
Which shows that if $v \in \mathrm{Null}(T-\lambda I)^j$, then $\overline{v} \in \mathrm{Null}(T-\overline{\lambda}I)^j$.
Apply the same process to $(T-\overline{\lambda}I)^j(\overline{v}) = 0$ to show that if $\overline{v} \in \mathrm{Null}(T-\overline{\lambda}I)^j$, then $v \in \mathrm{Null}(T-\lambda I)^j$.
Then it is easy to argue they have the same dimension, and from there you can go forth to make the conclusions about the Jordan blocks.
