Let $A$ be an $n\times n$ matrix with complex entries. Prove that if every characteristic value of $A$ is real, then $A$ is similar to a matrix with real entries.
My attempt: Suppose $A\in M_n(\Bbb{C})$. Let $f$ and $m$ be characteristic and minimal polynomial of $A$. Since $\Bbb{C}$ is algebraically closed and $f\in \Bbb{C}[x]$, we have $f$ splits into linear factors. That is $f=b(x-c_1)^{d_1}\cdots (x-c_k)^{d_k}$, where $b,c_i\in \Bbb{C}$ and $d_i\geq 1$. Since $f$ is a monic polynomial, $b=1$. So $c_1,…,c_k$ are eigenvalues of $A$. By hypothesis, $c_i\in \Bbb{R}$, for all $i\in J_k$. Since $f$ and $m$ have same roots, we have $m=(x-c_1)^{e_1}\cdots (x-c_k)^{e_k}$, where $e_i\geq 1$. So $m\in \Bbb{R}[x]$. Define $T:\Bbb{C}^n\to \Bbb{C}^n$ such that $[T]_B=A$, where $B$ is canonical basis of $\Bbb{C}^n$. By theorem 3 section 7.2, $\exists \alpha_1,…,\alpha_r\in \Bbb{C}^n\setminus \{0\}$ with respective $T$-annihilators $p_1,…,p_r$ such that $V=Z(\alpha_1;T)\oplus \cdots Z(\alpha_r;T)$, and $p_{i+1}|p_i$, and $p_1=m$. Since $p_i|p_1=m$, we have $p_i=(x-c_1)^{g_1}\cdots (x-c_k)^{g_k}$, where $0\leq g_i\leq e_i$. So $p_i\in \Bbb{R}[x]$, $\forall i\in J_r$. Then $\exists \mathcal{B}$ basis of $\Bbb{C}^n$ such that $$[T]_{\mathcal{B}}=\begin{bmatrix}A_1& & \\ & \ddots & \\ & & A_r \\ \end{bmatrix}$$ where $A_i$ is $k_i \times k_i$ companion matrix of $p_i$. So $A_i\in M_{k_i\times k_i}(\Bbb{R})$, $\forall i\in J_r$. Thus $[T]_\mathcal{B}\in M_n(\Bbb{R})$. Hence $A=[T]_B$ is similar over $\Bbb{C}$ to $[T]_\mathcal{B}$. Is my proof correct?
It is correct but could be shorter: by theorem 5 of the same section, $A$ is similar over $\Bbb C$ to a matrix $$\begin{bmatrix}A_1& & \\ & \ddots & \\ & & A_r \\ \end{bmatrix}$$ where each $A_i$ is the companion matrix of some monic polynomial $p_i.$ Since the roots of the characteristic polynomial $\prod p_i$ are real, so are the coefficients of each $p_i$ hence the entries of $A_i.$