Is my attempt at the proof of the following theorem Correct?
SOME PRELIMINARY NOTATION:
Given some choice of basis for a finite dimensional vector space $V$
- $\mathcal{M}(T)$ denotes the matrix of a linear transformation $T$ on that that vector space.
- $\mathcal{M}(v)$ denotes the matrix of the column vector of the vector $v$ under that basis.
Theorem. Given that $V$ is a complex vector space, $T\in\mathcal{L}(V)$ and the matrix of $T$ with respect to some basis of $V$ contains only real entries. Then if $\lambda$ is an eigenvalue of $T$, then so is $\overline\lambda$.
Proof. Let us first define the conjugate of a general $m\times n$ matrix A as follows. $$B\equiv\overline A\ \operatorname{iff}\ \forall i\in I_m = \{1,2,...,m\}\forall j\in I_n = \{1,2,...,n\}(B_{ij} = \overline A_{ij})$$ Assume now that $\dim V = n$ and that $\lambda$ is an eigenvalue of $T$ thus for some non-zero $v\in V$, $Tv = \lambda v$, equivalently in matrix form we have. $\mathcal{M}(T)\cdot\mathcal{M}(v) = \lambda\cdot\mathcal{M}(V)$ which when written out as a system of equations we have. $$\sum_{j=1}^{n}\mathcal{M}(T)_{ij}\cdot \mathcal{M}(v)_{j1} = \lambda \mathcal{M}(v)_{i1}\ \operatorname{for}\ \operatorname{all}\ i\in I_n$$ taking the conjugate of both sides of all of the above equations we have $$\overline{\sum_{j=1}^{n}\mathcal{M}(T)_{ij}\cdot \mathcal{M}(v)_{j1}} = \overline{\lambda\cdot\mathcal{M}(v)_{i1}}\ \operatorname{for}\ \operatorname{all}\ i\in I_n$$ and then by appealing to the additivity and multiplicativity of complex conjugates we have the following system. $$\sum_{j=1}^{n}\overline{\mathcal{M}(T)_{ij}}\cdot\overline{\mathcal{M}(v)_{j1}} = \overline{\lambda}\cdot\overline{\mathcal{M}(v)_{i1}}\ \operatorname{for}\ \operatorname{all}\ i\in I_n$$ which when written as a matrix equation yields $\overline{\mathcal{M}(T)}\cdot\overline{\mathcal{M}(v)} = \overline{\lambda}\cdot\overline{\mathcal{M}(v)}$ but $\overline{\mathcal{M}(T)} = \mathcal{M}(T)$ since all entries of $\mathcal{M}(T)$ are real consequently $\mathcal{M}(T)\cdot\overline{\mathcal{M}(v)} = \overline{\lambda}\cdot\overline{\mathcal{M}(v)}$ which implies that $\overline\lambda$ is an eigenvalue of $T$.
$\blacksquare$
Yes, it is correct.
You could also observe that, since your matrix is real, then its characteristic polynomial $P_T(x)$ only has real coefficients and therefore $P(\lambda)=0\implies P\left(\overline\lambda\right)=0$.