The well-known Pole Assignment Theorem is
Pole Assignment Theorem. Given $A\in \mathbb{R}^{n\times n},B\in \mathbb{R}^{n\times m}$, for any target pole set $P=\{\lambda_1,...,\lambda_n\}$ which is closed under complex conjugate, there exists a matrix $F\in \mathbb{R}^{m\times n}$ such that $\lambda(A+BF)=P$ if and only if $(A,B)$ is controllable.
If $\mu$ is an eigenvalue of $A$ and $\mu$ is still an eigenvalue of $A+BF$ for any $F\in \mathbb{R}^{m\times n}$, $\mu$ is called an uncontrollable pole of $A$, my question is
Question. The pole assignment problem has a solution i.e. given a target pole set $P$, there exists a matrix $F\in \mathbb{R}^{m\times n}$ such that $$\lambda(A+BF)=P$$ if and only if all uncontrollable poles of $A$ are in the target pole set $P$.
by real Schur decomposition, we can choose an orthogonal matrix $Q$ such that $$Q^TAQ=\begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}$$ where $\lambda(A_{11})$ consists of all the uncontrollable poles of $A$. Let $$Q^TB=\begin{bmatrix} B_1 \\B_2 \end{bmatrix}$$ but I do not have idea that how to continue... any help will be appreciated.
This is closely related to the Kalman decomposition, from which it can be noted that $\lambda(A_{22})$ instead of $\lambda(A_{11})$ should contain all uncontrollable poles of $A$. You want to show that the eigenvalues of $A + B\,F$ and thus also $Q^\top(A + B\,F)\,Q$ always contain the uncontrollable poles of $A$. Using this what can you say about $B_1$ and $B_2$? Hint: use $F\,Q = \begin{bmatrix}F_1 & F_2\end{bmatrix}$ and look at the eigenvalues of $Q^\top(A + B\,F)\,Q$.