[proofplan] We use the Kalman reachable subspace to separate the dynamics into a controllable block and an uncontrollable block. In that decomposition, feedback can move the controllable eigenvalues by pole placement, while the uncontrollable eigenvalues are fixed. The PBH rank condition is then shown to be exactly the statement that all fixed uncontrollable eigenvalues already lie in the open left half-plane. [/proofplan]

[step:Separate the reachable and unreachable dynamics by a Kalman decomposition]A real feedback matrix means a matrix $K \in \mathbb R^{m \times n}$, and the pair $(A,B)$ is stabilisable precisely when some such $K$ makes every eigenvalue of $A+BK$ have negative real part. Let $\mathcal R \subset \mathbb R^n$ denote the reachable subspace of the pair $(A,B)$, defined by \begin{align*} \mathcal R := \operatorname{span}_{\mathbb R}\{A^j Bv : 0 \le j \le n-1,\ v \in \mathbb R^m\}. \end{align*} Since $A(A^jBv)=A^{j+1}Bv$ for $0 \le j \le n-2$, and $A^nBv$ is a linear combination of $Bv,ABv,\dots,A^{n-1}Bv$ by the [Cayley-Hamilton Theorem](/theorems/865) applied to $A$, the subspace $\mathcal R$ is $A$-invariant. Let $r := \dim_{\mathbb R}\mathcal R$. Choose a basis $e_1,\dots,e_r$ of $\mathcal R$ and extend it to a basis $e_1,\dots,e_n$ of $\mathbb R^n$. Let $T \in \mathbb R^{n \times n}$ be the invertible matrix whose columns are this basis. Because $\mathcal R$ is $A$-invariant and $\operatorname{Range} B \subset \mathcal R$, there exist real matrices $A_{11} \in \mathbb R^{r \times r}$, $A_{12} \in \mathbb R^{r \times (n-r)}$, $A_{22} \in \mathbb R^{(n-r) \times (n-r)}$, and $B_1 \in \mathbb R^{r \times m}$ with the following block description relative to $\mathbb R^n = \mathbb R^r \oplus \mathbb R^{n-r}$: the upper-left, upper-right, lower-left, and lower-right blocks of $T^{-1}AT$ are respectively $A_{11}$, $A_{12}$, $0$, and $A_{22}$, while the upper and lower blocks of $T^{-1}B$ are respectively $B_1$ and $0$. For $0 \le j \le n-1$ and $v \in \mathbb R^m$, the vector $A^jBv$ has reduced coordinates $(A_{11}^jB_1v,0)$ in this basis. Since the first $r$ basis vectors span $\mathcal R$, the vectors $A_{11}^jB_1v$ with $0 \le j \le n-1$ and $v \in \mathbb R^m$ span $\mathbb R^r$. Applying the [Cayley-Hamilton Theorem](/theorems/865) to $A_{11} \in \mathbb R^{r \times r}$, every power $A_{11}^j$ with $j \ge r$ is a real linear combination of $I_r,A_{11},\dots,A_{11}^{r-1}$. Hence the same span is obtained using only $0 \le j \le r-1$, so the reachable subspace of $(A_{11},B_1)$ is all of $\mathbb R^r$. Thus the pair $(A_{11},B_1)$ is controllable on $\mathbb R^r$.[/step]

[guided]The reachable subspace is the part of the state space that can be affected by the input through repeated interaction with the drift matrix $A$. We define it explicitly by \begin{align*} \mathcal R := \operatorname{span}_{\mathbb R}\{A^j Bv : 0 \le j \le n-1,\ v \in \mathbb R^m\}. \end{align*} This is a real linear subspace of $\mathbb R^n$. It contains $\operatorname{Range}B$ because the term $j=0$ gives $Bv$. We next verify that $\mathcal R$ is invariant under $A$. If $0 \le j \le n-2$, then \begin{align*} A(A^jBv)=A^{j+1}Bv \in \mathcal R. \end{align*} For the remaining power, the [Cayley-Hamilton Theorem](/theorems/865) applied to $A$ expresses $A^n$ as a real linear combination of $I_n,A,\dots,A^{n-1}$. Hence $A^nBv$ is a real linear combination of $Bv,ABv,\dots,A^{n-1}Bv$, so it also belongs to $\mathcal R$. Therefore $A\mathcal R \subset \mathcal R$. Let $r:=\dim_{\mathbb R}\mathcal R$. Choose a basis $e_1,\dots,e_r$ of $\mathcal R$ and extend it to a basis $e_1,\dots,e_n$ of $\mathbb R^n$. Let $T \in \mathbb R^{n \times n}$ be the change-of-basis matrix with these columns. In this basis, the first $r$ coordinates describe $\mathcal R$. Since $A\mathcal R \subset \mathcal R$, the lower-left block of $T^{-1}AT$ is zero. Since $\operatorname{Range}B \subset \mathcal R$, the lower block of $T^{-1}B$ is zero. Thus there are real matrices $A_{11} \in \mathbb R^{r \times r}$, $A_{12} \in \mathbb R^{r \times (n-r)}$, $A_{22} \in \mathbb R^{(n-r) \times (n-r)}$, and $B_1 \in \mathbb R^{r \times m}$ with the following block description relative to $\mathbb R^n = \mathbb R^r \oplus \mathbb R^{n-r}$: the upper-left, upper-right, lower-left, and lower-right blocks of $T^{-1}AT$ are respectively $A_{11}$, $A_{12}$, $0$, and $A_{22}$, while the upper and lower blocks of $T^{-1}B$ are respectively $B_1$ and $0$. The point of this decomposition is that the lower block is not directly actuated. The upper block is precisely the reachable part: for each $0 \le j \le n-1$ and $v \in \mathbb R^m$, the vector $A^jBv$ has reduced coordinates $(A_{11}^jB_1v,0)$. Because $e_1,\dots,e_r$ form a basis of $\mathcal R$, the span of the vectors $A_{11}^jB_1v$ with $0 \le j \le n-1$ is all of $\mathbb R^r$. To match the usual controllability criterion for the $r$-dimensional pair, we reduce the powers. Applying the [Cayley-Hamilton Theorem](/theorems/865) to $A_{11}$ expresses every $A_{11}^j$ with $j \ge r$ as a real linear combination of $I_r,A_{11},\dots,A_{11}^{r-1}$. Therefore the vectors $A_{11}^jB_1v$ with $0 \le j \le r-1$ and $v \in \mathbb R^m$ still span $\mathbb R^r$. Hence the reachable subspace of $(A_{11},B_1)$ is $\mathbb R^r$, so $(A_{11},B_1)$ is controllable on $\mathbb R^r$.[/guided]

[step:Identify which eigenvalues feedback cannot move] Let $K \in \mathbb R^{m \times n}$ be arbitrary, and write $KT=(K_1\ K_2)$ with $K_1 \in \mathbb R^{m \times r}$ and $K_2 \in \mathbb R^{m \times (n-r)}$. The matrix $T^{-1}(A+BK)T$ is block upper triangular relative to $\mathbb R^n = \mathbb R^r \oplus \mathbb R^{n-r}$: its upper-left, upper-right, lower-left, and lower-right blocks are respectively $A_{11}+B_1K_1$, $A_{12}+B_1K_2$, $0$, and $A_{22}$. Therefore its characteristic polynomial factors as \begin{align*} \det(\mu I_n-T^{-1}(A+BK)T) = \det(\mu I_r-(A_{11}+B_1K_1))\det(\mu I_{n-r}-A_{22}) \end{align*} for every $\mu \in \mathbb C$. Hence the eigenvalues of $A_{22}$ are eigenvalues of $A+BK$ for every feedback matrix $K$. Conversely, the hypotheses of the [Pole Placement Theorem](/theorems/3947) are satisfied because $(A_{11},B_1)$ is a real controllable pair on $\mathbb R^r$. That theorem applies to any multiset of $r$ complex numbers counted with multiplicity and closed under complex conjugation. Choose the concrete multiset $\{-1,\dots,-1\}$, where $-1$ is repeated $r$ times; it is contained in the open left half-plane and is closed under complex conjugation. The theorem gives a real matrix $K_1 \in \mathbb R^{m \times r}$ such that every eigenvalue of $A_{11}+B_1K_1$ equals $-1$ and hence has negative real part. Taking $K_2=0$ and defining $K \in \mathbb R^{m \times n}$ by $KT=(K_1\ 0)$, the eigenvalues of $A+BK$ are exactly the union of the eigenvalues of $A_{11}+B_1K_1$ and $A_{22}$, counted with algebraic multiplicity. Therefore $(A,B)$ is stabilisable if and only if every eigenvalue of $A_{22}$ has negative real part. [/step]

[step:Translate the PBH rank condition into a condition on the unreachable block]Fix $\lambda \in \mathbb C$. Define the complex [linear map](/page/Linear%20Map) $N_\lambda: \mathbb C^r \times \mathbb C^{n-r} \times \mathbb C^m \to \mathbb C^r \times \mathbb C^{n-r}$ by \begin{align*} N_\lambda(x_1,x_2,u)=((\lambda I_r-A_{11})x_1-A_{12}x_2+B_1u,(\lambda I_{n-r}-A_{22})x_2). \end{align*} The matrix of $N_\lambda$ is obtained from $\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}$ by left multiplication by $T^{-1}$ and by multiplying the first $n$ columns on the right by $T$, equivalently by the invertible block column operation $\operatorname{diag}(T,I_m)$. Since both operations are invertible over $\mathbb C$, they preserve rank, and therefore \begin{align*} \operatorname{rank}_{\mathbb C}\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}=\operatorname{rank}_{\mathbb C}N_\lambda. \end{align*} We claim that \begin{align*} \operatorname{rank}_{\mathbb C}\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}<n \end{align*} if and only if $\lambda$ is an eigenvalue of $A_{22}$. Indeed, if $\lambda I_{n-r}-A_{22}$ is singular, then its left nullspace is nonzero, so choose $q_2 \in \mathbb C^{n-r}$ with $q_2 \neq 0$ and \begin{align*} q_2^*(\lambda I_{n-r}-A_{22})=0. \end{align*} The nonzero row functional $q^*: \mathbb C^r \times \mathbb C^{n-r} \to \mathbb C$ defined by $q^*(y_1,y_2)=q_2^*y_2$ annihilates $N_\lambda$, so $\operatorname{rank}_{\mathbb C}N_\lambda<n$. Conversely, suppose $\operatorname{rank}_{\mathbb C}N_\lambda<n$. Then there exists a nonzero row functional $q^*: \mathbb C^r \times \mathbb C^{n-r} \to \mathbb C$ of the form $q^*(y_1,y_2)=q_1^*y_1+q_2^*y_2$, with $q_1 \in \mathbb C^r$ and $q_2 \in \mathbb C^{n-r}$, such that $q^*N_\lambda=0$. Evaluating this identity on triples $(x_1,0,0)$ and $(0,0,u)$ gives \begin{align*} q_1^*(\lambda I_r-A_{11})=0, \qquad q_1^*B_1=0. \end{align*} We now use the controllability of $(A_{11},B_1)$ directly. If $q_1 \neq 0$, then $q_1^*(\lambda I_r-A_{11})=0$ gives $q_1^*A_{11}=\lambda q_1^*$, and hence $q_1^*A_{11}^j=\lambda^j q_1^*$ for every integer $j \ge 0$. Together with $q_1^*B_1=0$, this implies $q_1^*A_{11}^jB_1v=0$ for every $0 \le j \le r-1$ and every $v \in \mathbb C^m$. These vectors span $\mathbb C^r$ because $(A_{11},B_1)$ is controllable, so $q_1^*$ vanishes on all of $\mathbb C^r$, contradicting $q_1 \neq 0$. Therefore $q_1=0$. Evaluating $q^*N_\lambda=0$ on triples $(0,x_2,0)$ then gives \begin{align*} q_2^*(\lambda I_{n-r}-A_{22})=0. \end{align*} Since $q^*\neq 0$ and $q_1=0$, we have $q_2\neq 0$, so $\lambda$ is an eigenvalue of $A_{22}$.[/step]

[guided]The rank comparison needs two invertible operations, not only left multiplication. We first left-multiply $\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}$ by $T^{-1}$, and then multiply the first $n$ columns by $T$. This second operation is the invertible block column operation $\operatorname{diag}(T,I_m)$. After both operations, the resulting complex linear map is $N_\lambda: \mathbb C^r \times \mathbb C^{n-r} \times \mathbb C^m \to \mathbb C^r \times \mathbb C^{n-r}$, defined by \begin{align*} N_\lambda(x_1,x_2,u)=((\lambda I_r-A_{11})x_1-A_{12}x_2+B_1u,(\lambda I_{n-r}-A_{22})x_2). \end{align*} Because the two operations are invertible over $\mathbb C$, they do not change rank, so \begin{align*} \operatorname{rank}_{\mathbb C}\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}=\operatorname{rank}_{\mathbb C}N_\lambda. \end{align*} We now identify exactly when $N_\lambda$ fails to have full row rank $n$. If $\lambda I_{n-r}-A_{22}$ is singular, its left nullspace is nonzero. Choose $q_2 \in \mathbb C^{n-r}$ with $q_2 \neq 0$ and \begin{align*} q_2^*(\lambda I_{n-r}-A_{22})=0. \end{align*} Then the row functional $q^*: \mathbb C^r \times \mathbb C^{n-r} \to \mathbb C$ given by $q^*(y_1,y_2)=q_2^*y_2$ annihilates every value of $N_\lambda$. Hence the row rank of $N_\lambda$ is less than $n$. For the converse, assume $\operatorname{rank}_{\mathbb C}N_\lambda<n$. Then some nonzero row functional annihilates its range. Write this functional as $q^*(y_1,y_2)=q_1^*y_1+q_2^*y_2$, where $q_1 \in \mathbb C^r$ and $q_2 \in \mathbb C^{n-r}$. The identity $q^*N_\lambda=0$ holds for every $x_1 \in \mathbb C^r$, $x_2 \in \mathbb C^{n-r}$, and $u \in \mathbb C^m$. Setting $x_2=0$ and $u=0$ gives \begin{align*} q_1^*(\lambda I_r-A_{11})=0. \end{align*} Setting $x_1=0$ and $x_2=0$ gives \begin{align*} q_1^*B_1=0. \end{align*} We do not need to cite a separate PBH controllability theorem here; the required implication follows from the controllability already proved for $(A_{11},B_1)$. Suppose, for contradiction, that $q_1 \neq 0$. From $q_1^*(\lambda I_r-A_{11})=0$ we get $q_1^*A_{11}=\lambda q_1^*$, so induction gives $q_1^*A_{11}^j=\lambda^j q_1^*$ for every integer $j \ge 0$. Since $q_1^*B_1=0$, it follows that $q_1^*A_{11}^jB_1v=0$ for every $0 \le j \le r-1$ and every $v \in \mathbb C^m$. But controllability means precisely that these vectors span $\mathbb C^r$. Thus $q_1^*$ vanishes on all of $\mathbb C^r$, contradicting $q_1 \neq 0$. Therefore $q_1=0$. With $q_1=0$, the identity $q^*N_\lambda=0$ evaluated at $(0,x_2,0)$ becomes \begin{align*} q_2^*(\lambda I_{n-r}-A_{22})=0. \end{align*} The functional $q^*$ is nonzero, and $q_1=0$, so $q_2\neq 0$. Thus $\lambda I_{n-r}-A_{22}$ is singular, meaning that $\lambda$ is an eigenvalue of $A_{22}$.[/guided]

[step:Conclude equivalence with stabilizability] By the previous step, the rank condition \begin{align*} \operatorname{rank}_{\mathbb C}\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}=n \end{align*} for every $\lambda \in \mathbb C$ with $\operatorname{Re}\lambda \ge 0$ is equivalent to saying that $A_{22}$ has no eigenvalue in the closed right half-plane. This is equivalent to every eigenvalue of $A_{22}$ having strictly negative real part. By the feedback-invariance analysis above, $(A,B)$ is stabilisable if and only if every eigenvalue of $A_{22}$ has strictly negative real part. Combining the two equivalences proves that $(A,B)$ is stabilisable if and only if \begin{align*} \operatorname{rank}_{\mathbb C}\begin{pmatrix}\lambda I_n-A&B\end{pmatrix}=n \end{align*} for every $\lambda \in \mathbb C$ satisfying $\operatorname{Re}\lambda \ge 0$. [/step]