Prove that $\dim G(\lambda,T_{\Bbb{C}})=\dim G(\bar{\lambda},T_{\Bbb{C}})$ Consider the following theorem in Linear Algebra Done Right. 
$V$ is a vector space and $T$ is an operator on $V$. Denote the complexification of $V$ and $T$, respectively, by $V_{\Bbb{C}}$ and $T_{\Bbb{C}}$. Also, in the book, multiplicity of an eigen-value is defined as $\dim G(\lambda,T)$ where $G(\lambda,T)$ is the generalized eigen-space. The theorem $9.17$ is trying to prove
$$\dim G(\lambda,T_{\Bbb{C}})=\dim G(\bar{\lambda},T_{\Bbb{C}})$$
I don't understand how routine arguments work here! I can understand that if $u+iv \in G(\lambda,T)$ then $u-iv \in G(\bar{\lambda},T)$ but the rest is not too much clear for me. Any hint is appreciated.



Linear independence: let $\alpha_1,\dots,\alpha_m\in\Bbb C$ s.t. $\displaystyle\sum_{k=1}^m \alpha_k(u_k-iv_k)=0$. We apply complex conjugation: $\displaystyle\sum_{k=1}^m \overline{\alpha_k}(u_k+iv_k)=0$. Since $(u_1+iv_1,\dots,u_m+iv_m)$ is linearly independent, we see that $\forall k\in\{1,\dots,m\},\overline{\alpha_k}=0$ thus $\forall k,\,\alpha_k=0$. 
Spanning: let $u_0+iv_0=u_0-i(-v_0)=u-iv\in G(\overline{\lambda},T_{\Bbb C})$. By definition, there is an integer $j$ s.t. $(T_{\Bbb C}-\bar{\lambda}I)^j(u-iv)=0$. By Lemma 9.12, we have $(T_{\Bbb C}-\lambda I)^j(u+iv)=0$. Thus $u+iv\in G(\lambda,T_{\Bbb C})$ and, since $(u_1+iv_1,\dots,u_m+iv_m)$ spans $G(\lambda,T_{\Bbb C})$, there exist $\alpha_1,\dots,\alpha_m\in\Bbb C$ s.t.  $\displaystyle\sum_{k=1}^m \alpha_k(u_k+iv_k)=u+iv$. By conjugating, we see that $\displaystyle\sum_{k=1}^m \overline{\alpha_k}(u_k-iv_k)=u-iv$, hence $(u_1-iv_1,\dots,u_m-iv_m)$ spans $G(\bar{\lambda},T_{\Bbb C})$. 
