Proof Verification: Eigenvalues of a Complex Vector Space I think I found a solution to question 16 from chapter 5.A of Axler's Linear Algebra book which states

Suppose V is a complex vector space and $T\in L(v)$, and the matrix of T with respect to some basis of V only has real entries show that if $\lambda$ is an eigenvalue then so is its complex conjugate.

and wanted to verify that the solution I attempted is indeed correct.
My Proof Attempt: 
First I claim that for this matrix with the particular basis $C\circ T\circ C = T$ where $C$ is the complex conjugate map. Given any vector $v=\sum_{k=1}^{n}b_kv_k$, where $v_k$ is the particular basis, $T\circ C= \sum_{k=1}^{n}\overline{b_k}T(v_k)$. Hence $C\circ T\circ C= \sum_{k=1}^{n}b_k\overline{T(v_k)}$. But since the entries are real $T(v_k)=\overline{T(V_k)}$. Therefore the first claim is as such. 
To prove the claim let $v$ be the given eigenvector of $T$ then $C\circ T\circ C (v)= \lambda v$, for some non-zero complex number $\lambda$. Since $C$ is its own inverse it then follows that $T\circ C(v) = C(\lambda v) $. By definition the map $C$ takes the coefficient of the vector and gives the vector with co-coefficients which are the complex conjugate of the orginal. Thus it is clear that $T\circ C(v) = \overline{\lambda}C(v)$ and the proof is complete. 
I was wondering if this is correct and is clear enough, any critques and suggestions would be greatly appreciated, thank you!
 A: The idea of the proof is correct but there is one technical problem that should be addressed: A complex vector space $V$ has no natural conjugation map $C \colon V \rightarrow V$ so this map should be defined explicitly. Similarly, the expression $\overline{T(v_k)}$ makes no sense because.
Let me suggest how to fix that. If $\beta = (v_1,\dots,v_n)$ is the basis of $V$ with respect to which the matrix of $T$ is real, we have an isomorphism $\Phi \colon \mathbb{C}^n \rightarrow V$ given by
$$ \Phi \left( \sum_{i=1}^n a_i e_i \right) = \sum_{i=1}^n a_i v_i $$
where $(e_1,\dots,e_n)$ is the standard basis of $\mathbb{C}^n$ and $a_i \in \mathbb{C}$. Define the operator $T' \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ by $T' := \Phi^{-1} \circ T \circ \Phi$. Then $T$ and $T'$ have the same eigenvalues and the matrix representing $T'$ with respect to the standard basis $(e_1,\dots,e_n)$ of $\mathbb{C}^n$ is the same as the matrix representing $T$ with respect to the basis $(v_1,\dots,v_n)$ of $V$ so it has real entries.
Thus, by replacing $T$ with $T'$ and $V$ with $\mathbb{C}^n$ we need to solve the following question:
Let $T \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ be an operator such that $T(e_i)$ is a real linear combination of $(e_1,\dots,e_n)$ for all $1 \leq i \leq n$. Show that if $\lambda \in \mathbb{C}$ is an eigenvalue of $T$ then $\overline{\lambda}$ is an eigenvalue of $T$.
Now, in $\mathbb{C}^n$ a vector looks like $v = (z_1,\dots,z_n)$ so we can talk about $\overline{v} = (\overline{z}_1,\dots,\overline{z}_n)$ and your proof carries more or less through. Alternatively, it is actually easier to translate everything to matrices and solve it on the level of matrices.
A: T has only real values => Any complex vector field is being mapped to a real vector field.
T:C^N->R^N
Any complex vector v can be written as a+bi where a & b are vectors with real components.
Tv=zv   : z is the eigenvalue here
Substituting v with a+bi & a-bi and z with x&y respectively we get:
T(a+bi)=Ta+iTb=Ta=x(a+bi) -> The complex portion is 0 [A]
T(a-bi)=Ta-iTb=Ta=y(a-bi) -> The complex portion is 0 [B]
*(above: iTb is 0 as Ta & Tb are real and iTb is complex; the map is from complex to real so the complex portion is 0)
Let x=c+di & y=e+fi
from [A]:
complex [(c+di)(a+bi)] =0
=> bc+ad=0  [C]
from [B]:
complex [(e+fi)(a-bi)]=0
=> af-be=0  [D]
Also, the real portions of x(a+bi) & y(a-bi) are equal since, T(a+bi)=Ta+iTb=Ta & T(a-bi)=Ta-iTb=Ta (complex portion 0)
=> Real [(c+di)(a+bi)] = Real [(e+fi)(a-bi)]
=> ac-bd=ae+bf  [E]
Solve B,D & E to prove c=e & d=-f
aX[E] => (a^2)c-abd=(a^2)e+abf   [F]
bX[C] => -abd=(b^2)c
bX[D] => abf=(b^2)e
Substituting in [F]
(a^2)c+(b^2)c=(a^2)e+(b^2)e
=> c=e
Using c=e in [C] & [D]
d=-f
=> x=conjugate(y)
