Let $A\in M_{n}(\mathbb{R})$ and $B\in\mathbb{R}^{n}$. Then the eigenvalues of $A+BF$ can be assigned arbitrarily if $(A,B)$ is Controllable. i.e. if $(A,B)$ controllable then $$\forall M\in M_{n}(\mathbb{R}),\exists F\in M_{1\times n}(\mathbb{R}):\mathcal{X}_{A+BF}(s)=\mathcal{X}_{M}(s)$$where $\mathcal{X}(\cdot)$ denotes the characteristic polynomial.

Let $A\in M_{n}(\mathbb{R})$ and $B\in M_{n\times m}(\mathbb{R})$. Then the eigenvalues of $A+BF$ can be assigned arbitrarily if $(A,B)$ is Controllable. i.e. if $(A,B)$ controllable then $$\forall M\in M_{n}(\mathbb{R}),\exists F\in M_{1\times n}(\mathbb{R}):\mathcal{X}_{A+BF}(s)=\mathcal{X}_{M}(s)$$where $\mathcal{X}(\cdot)$ denotes the characteristic polynomial.

Let $A\in M_{n}(\mathbb{R}),B\in M_{n\times m} (\mathbb{R})$ and suppose $(A,B)$ is not Controllable, then $$\exists M\in M_{n}(\mathbb{R}):\forall F\in M_{m\times n}(\mathbb{R})\,\mathcal{X}_{A+BF}(s)\not=\mathcal{X}_{M}(s)$$