[proofplan] We pass to the Kalman controllability decomposition, which separates the state space into the controllable part and an uncontrollable quotient. In these coordinates, every static state feedback changes only the controllable diagonal block and the upper-right coupling, while the uncontrollable diagonal block remains unchanged. Therefore the uncontrollable eigenvalues are unavoidable closed-loop eigenvalues, and stabilisation is possible exactly when those unavoidable eigenvalues already lie in the open left half-plane and the controllable block is stabilised by pole placement. [/proofplan]

[step:Put the system into Kalman controllability form]Let $\mathcal{C} \subset \mathbb{R}^n$ denote the controllable subspace of the pair $(A,B)$, defined by \begin{align*} \mathcal{C} := \operatorname{span}_{\mathbb{R}}\{A^j Bv : 0 \leq j \leq n-1,\ v \in \mathbb{R}^m\}. \end{align*} Let $r := \dim \mathcal{C}$. Since $A\mathcal{C} \subset \mathcal{C}$ and $\operatorname{Range}(B) \subset \mathcal{C}$, choose a basis of $\mathbb{R}^n$ whose first $r$ vectors form a basis of $\mathcal{C}$, and let $T \in GL(n,\mathbb{R})$ be the change-of-basis matrix with those basis vectors as columns. In the coordinates $z = T^{-1}x$, define \begin{align*} \widetilde{A} := T^{-1}AT \in \mathbb{R}^{n \times n} \end{align*} and \begin{align*} \widetilde{B} := T^{-1}B \in \mathbb{R}^{n \times m}. \end{align*} The invariance $A\mathcal{C} \subset \mathcal{C}$ and the inclusion $\operatorname{Range}(B) \subset \mathcal{C}$ imply that there are matrices \begin{align*} A_c \in \mathbb{R}^{r \times r}, \quad A_{12} \in \mathbb{R}^{r \times (n-r)}, \quad A_u \in \mathbb{R}^{(n-r) \times (n-r)}, \quad B_c \in \mathbb{R}^{r \times m} \end{align*} such that, with respect to the decomposition $\mathbb{R}^n = \mathbb{R}^r \oplus \mathbb{R}^{n-r}$, the transformed dynamics are given by \begin{align*} \widetilde{A}(z_c,z_u) = (A_c z_c + A_{12}z_u, A_u z_u) \end{align*} for $z_c \in \mathbb{R}^r$ and $z_u \in \mathbb{R}^{n-r}$, and the transformed input map is given by \begin{align*} \widetilde{B}v = (B_c v,0) \end{align*} for $v \in \mathbb{R}^m$. These formulas include the degenerate cases $r=0$ and $r=n$, with the corresponding zero-dimensional summand interpreted in the standard way. Moreover, by construction, the pair $(A_c,B_c)$ is controllable, and the eigenvalues of $A_u$ are precisely the uncontrollable eigenvalues of $(A,B)$ in the Popov-Belevitch-Hautus sense: \begin{align*} \operatorname{rank}_{\mathbb{C}}\begin{pmatrix}\lambda I_n - A & B\end{pmatrix} < n. \end{align*} This is the Kalman controllability decomposition. The theorem is being cited as an external standard result not yet resolved in the wiki: Kalman controllability decomposition.[/step]

[guided]The purpose of this step is to choose coordinates that make the controllable and uncontrollable directions visible. Define the controllable subspace \begin{align*} \mathcal{C} := \operatorname{span}_{\mathbb{R}}\{A^j Bv : 0 \leq j \leq n-1,\ v \in \mathbb{R}^m\}. \end{align*} This is the smallest $A$-invariant subspace containing $\operatorname{Range}(B)$: the inclusion $\operatorname{Range}(B) \subset \mathcal{C}$ follows from the term $j=0$, and $A\mathcal{C} \subset \mathcal{C}$ follows from the Cayley-Hamilton reduction of powers $A^n,A^{n+1},\dots$ to linear combinations of $I_n,A,\dots,A^{n-1}$. Let $r := \dim \mathcal{C}$. Choose a basis of $\mathbb{R}^n$ whose first $r$ vectors form a basis of $\mathcal{C}$, and let $T \in GL(n,\mathbb{R})$ be the matrix whose columns are this basis. Define the transformed matrices \begin{align*} \widetilde{A} := T^{-1}AT \in \mathbb{R}^{n \times n} \end{align*} and \begin{align*} \widetilde{B} := T^{-1}B \in \mathbb{R}^{n \times m}. \end{align*} Because $A$ maps $\mathcal{C}$ into itself, the lower-left block of $\widetilde{A}$ is zero. Because the range of $B$ is contained in $\mathcal{C}$, the lower block of $\widetilde{B}$ is zero. Therefore there are matrices \begin{align*} A_c \in \mathbb{R}^{r \times r}, \quad A_{12} \in \mathbb{R}^{r \times (n-r)}, \quad A_u \in \mathbb{R}^{(n-r) \times (n-r)}, \quad B_c \in \mathbb{R}^{r \times m} \end{align*} with the following action on $\mathbb{R}^r \oplus \mathbb{R}^{n-r}$: \begin{align*} \widetilde{A}(z_c,z_u) = (A_c z_c + A_{12}z_u, A_u z_u) \end{align*} for $z_c \in \mathbb{R}^r$ and $z_u \in \mathbb{R}^{n-r}$, and \begin{align*} \widetilde{B}v = (B_c v,0) \end{align*} for $v \in \mathbb{R}^m$. If $r=0$ or $r=n$, the same notation is read with the evident zero-dimensional summand omitted. The Kalman controllability decomposition says more than just this block form: it also says that $(A_c,B_c)$ is controllable and that the spectrum of $A_u$ is exactly the set of uncontrollable eigenvalues of $(A,B)$, equivalently the eigenvalues $\lambda \in \mathbb{C}$ for which \begin{align*} \operatorname{rank}_{\mathbb{C}}\begin{pmatrix}\lambda I_n - A & B\end{pmatrix} < n. \end{align*} This result is being used as an external standard result not yet resolved in the wiki: Kalman controllability decomposition.[/guided]

[step:Show that feedback cannot move the uncontrollable block] Let $K \in \mathbb{R}^{m \times n}$ be an arbitrary state-feedback matrix. Define the transformed feedback matrix \begin{align*} \widetilde{K} := KT \in \mathbb{R}^{m \times n}. \end{align*} Write $\widetilde{K}$ in block form as \begin{align*} \widetilde{K} = \begin{pmatrix} K_c & K_u \end{pmatrix}, \end{align*} where $K_c \in \mathbb{R}^{m \times r}$ and $K_u \in \mathbb{R}^{m \times (n-r)}$. Since similarity preserves eigenvalues, \begin{align*} A + BK \text{ is Hurwitz} \end{align*} if and only if \begin{align*} T^{-1}(A+BK)T \end{align*} is Hurwitz. Compute \begin{align*} T^{-1}(A+BK)T = \widetilde{A} + \widetilde{B}\widetilde{K}. \end{align*} Using the block action above, for $z_c \in \mathbb{R}^r$ and $z_u \in \mathbb{R}^{n-r}$ we have \begin{align*} (\widetilde{A} + \widetilde{B}\widetilde{K})(z_c,z_u) = ((A_c + B_cK_c)z_c + (A_{12} + B_cK_u)z_u, A_u z_u). \end{align*} Thus $\widetilde{A} + \widetilde{B}\widetilde{K}$ is block upper triangular with diagonal blocks $A_c+B_cK_c$ and $A_u$, so its characteristic polynomial factors as \begin{align*} \det(\lambda I_n - (\widetilde{A}+\widetilde{B}\widetilde{K})) = \det(\lambda I_r - (A_c+B_cK_c))\det(\lambda I_{n-r}-A_u). \end{align*} Hence every eigenvalue of $A_u$ is an eigenvalue of every closed-loop matrix $A+BK$. [/step]

[step:Stabilise the controllable block when all uncontrollable eigenvalues are stable] Assume every uncontrollable eigenvalue of $(A,B)$ has negative real part. By the Kalman decomposition, every eigenvalue of $A_u$ has negative real part, so $A_u$ is Hurwitz. Since $(A_c,B_c)$ is controllable, the [pole placement theorem](/theorems/6395) for controllable finite-dimensional linear systems gives a matrix $K_c \in \mathbb{R}^{m \times r}$ such that $A_c + B_cK_c$ is Hurwitz. This invokes an external standard result not yet resolved in the wiki: pole placement theorem for controllable pairs. Choose \begin{align*} K_u := 0 \in \mathbb{R}^{m \times (n-r)} \end{align*} and define \begin{align*} \widetilde{K} := \begin{pmatrix} K_c & K_u \end{pmatrix} \in \mathbb{R}^{m \times n}. \end{align*} Set \begin{align*} K := \widetilde{K}T^{-1} \in \mathbb{R}^{m \times n}. \end{align*} Then $\widetilde{K} = KT$, and the transformed closed-loop operator acts by \begin{align*} T^{-1}(A+BK)T(z_c,z_u) = ((A_c + B_cK_c)z_c + A_{12}z_u, A_u z_u) \end{align*} for $z_c \in \mathbb{R}^r$ and $z_u \in \mathbb{R}^{n-r}$. Its spectrum is the union of the spectra of the two diagonal blocks. Both $A_c+B_cK_c$ and $A_u$ are Hurwitz, so $T^{-1}(A+BK)T$ is Hurwitz. Since Hurwitzness is invariant under similarity, $A+BK$ is Hurwitz. Thus $(A,B)$ is stabilisable. [/step]

[step:Rule out stabilisation when an uncontrollable eigenvalue is unstable] Conversely, assume $(A,B)$ is stabilisable. Then there exists $K \in \mathbb{R}^{m \times n}$ such that $A+BK$ is Hurwitz. Let $\lambda \in \mathbb{C}$ be an uncontrollable eigenvalue of $(A,B)$. By the Kalman decomposition, $\lambda$ is an eigenvalue of $A_u$. By the block-triangular computation above, $\lambda$ is therefore an eigenvalue of $T^{-1}(A+BK)T$, and hence also an eigenvalue of $A+BK$. Since $A+BK$ is Hurwitz, every eigenvalue of $A+BK$ has negative real part. Therefore \begin{align*} \operatorname{Re}\lambda < 0. \end{align*} Thus every uncontrollable eigenvalue of $(A,B)$ has negative real part. [/step]

[step:Combine the two implications] The preceding step proves that stabilisability forces every uncontrollable eigenvalue to lie in the open left half-plane. The construction using pole placement proves the converse: if every uncontrollable eigenvalue lies in the open left half-plane, then there is a real feedback matrix $K \in \mathbb{R}^{m \times n}$ such that $A+BK$ is Hurwitz. Hence $(A,B)$ is stabilisable if and only if every uncontrollable eigenvalue of $A$ for the pair $(A,B)$ has negative real part. [/step]