Arbitrary assignment of Eigenvalues

Theorem ((Single-input) Arbitrary assignment of Eigenvalues)

Let A∈Mn(R)A\in M_{n}(\mathbb{R}) and B∈RnB\in\mathbb{R}^{n}. Then the eigenvalues of A+BFA+BF can be assigned arbitrarily if (A,B)(A,B) is Controllable. i.e. if (A,B)(A,B) controllable then ∀M∈Mn(R),∃F∈M1×n(R):XA+BF(s)=XM(s)\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 X(⋅)\mathcal{X}(\cdot) denotes the characteristic polynomial.

Theorem ((Multi-input) Arbitrary assignment of Eigenvalues)

Let A∈Mn(R)A\in M_{n}(\mathbb{R}) and B∈Mn×m(R)B\in M_{n\times m}(\mathbb{R}). Then the eigenvalues of A+BFA+BF can be assigned arbitrarily if (A,B)(A,B) is Controllable. i.e. if (A,B)(A,B) controllable then ∀M∈Mn(R),∃F∈M1×n(R):XA+BF(s)=XM(s)\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 X(⋅)\mathcal{X}(\cdot) denotes the characteristic polynomial.

Theorem (Lack of Controllability implies we cannot arbitrarily assign eigenvalues)

Let A∈Mn(R),B∈Mn×m(R)A\in M_{n}(\mathbb{R}),B\in M_{n\times m} (\mathbb{R}) and suppose (A,B)(A,B) is not Controllable, then ∃M∈Mn(R):∀F∈Mm×n(R) XA+BF(s)≠XM(s)\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)