[proofplan] We identify the rank condition with the absence of an eigenvector of $A$ at $\lambda$ that is invisible to the output matrix $C$. The unobservable subspace $\mathcal{N}$ is exactly the space of states whose entire output trajectory under the autonomous dynamics is zero, and it is invariant under $A$. Therefore an invisible eigenvector for $A$ is the same thing as an eigenvector of the restricted operator $A|_{\mathcal{N}}$. The rank condition over the closed right half-plane is thus equivalent to saying that the spectrum of $A|_{\mathcal{N}}$ lies in the open left half-plane, which is precisely detectability. [/proofplan] [step:Show that the unobservable subspace is invariant under $A$] Let $n$ and $m$ be positive integers denoting the matrix dimensions, so $A \in \mathbb{C}^{n \times n}$ and $C \in \mathbb{C}^{m \times n}$. Let $I_n \in \mathbb{C}^{n \times n}$ denote the identity matrix. Define the unobservable subspace $\mathcal{N} \subseteq \mathbb{C}^n$ by $\mathcal{N}:=\bigcap_{j=0}^{n-1}\ker(CA^j)$. Define the [linear map](/page/Linear%20Map) $B:\mathcal{N} \to \mathcal{N}$ by $B(x)=Ax$. We first verify that this map is well-defined. Let $x \in \mathcal{N}$. For each integer $k$ with $0 \le k \le n-2$, \begin{align*} CA^k(Ax)=CA^{k+1}x=0, \end{align*} because $x \in \ker(CA^{k+1})$. It remains to verify the condition for $k=n-1$. Let $p_A(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_1t+a_0$ be the characteristic polynomial of $A$. Since $A \in \mathbb{C}^{n \times n}$ is a finite-dimensional square matrix by the dimension declaration above, the [Cayley-Hamilton theorem](/theorems/865) applies and gives $A^n=-a_{n-1}A^{n-1}-\cdots-a_1A-a_0I_n$. Multiplying by $C$ on the left and evaluating at $x$ gives \begin{align*} CA^n x=-a_{n-1}CA^{n-1}x-\cdots-a_1CAx-a_0Cx=0. \end{align*} Therefore $CA^k(Ax)=0$ for every $0 \le k \le n-1$, so $Ax \in \mathcal{N}$. Hence $B=A|_{\mathcal{N}}$ is a well-defined linear operator. [/step] [step:Translate rank failure into an invisible eigenvector] Fix $\lambda \in \mathbb{C}$. Since $C \in \mathbb{C}^{m \times n}$, the integer $m$ is the output dimension. Let $M(\lambda):\mathbb{C}^n\to\mathbb{C}^{n+m}$ denote the linear map whose first $n$ output coordinates are $(\lambda I_n-A)v$ and whose final $m$ output coordinates are $Cv$ for $v \in \mathbb{C}^n$. This is the linear map represented by the matrix obtained by vertically stacking $\lambda I_n-A$ above $C$. The domain of $M(\lambda)$ is the $n$-dimensional [vector space](/page/Vector%20Space) $\mathbb{C}^n$. A linear map from an $n$-dimensional vector space has rank $n$ exactly when its kernel is $\{0\}$: rank $n$ means that the image has the same dimension as the domain, which is equivalent in finite dimension to injectivity. Hence $M(\lambda)$ has rank strictly smaller than $n$ if and only if its kernel is nonzero. Thus the rank condition fails at $\lambda$ if and only if there exists $v \in \mathbb{C}^n\setminus\{0\}$ such that $(\lambda I_n-A)v=0$ and $Cv=0$. Equivalently, $Av=\lambda v$ and $v$ is invisible to the output matrix $C$. [guided] Because $C \in \mathbb{C}^{m \times n}$, the integer $m$ is the number of output coordinates. The stacked matrix $M(\lambda)$ is the linear map $M(\lambda):\mathbb{C}^n\to\mathbb{C}^{n+m}$ whose first $n$ coordinates are $(\lambda I_n-A)v$ and whose last $m$ coordinates are $Cv$. The domain of $M(\lambda)$ has dimension $n$. If $\ker M(\lambda)=\{0\}$, then the images under $M(\lambda)$ of any basis of $\mathbb{C}^n$ are linearly independent, so the image has dimension $n$ and $\operatorname{rank}M(\lambda)=n$. Conversely, if $\operatorname{rank}M(\lambda)=n$, then the image has the same dimension as the domain, so $M(\lambda)$ is injective and $\ker M(\lambda)=\{0\}$. Therefore $M(\lambda)$ has full column rank, meaning rank $n$, exactly when its kernel is only $\{0\}$. Consequently rank failure is exactly the existence of a nonzero vector $v \in \mathbb{C}^n$ satisfying $M(\lambda)v=0$. Writing out the two block rows of this equation gives $(\lambda I_n-A)v=0$ and $Cv=0$. The first equation is precisely the eigenvector equation $Av=\lambda v$. The second equation says that the output map $C:\mathbb{C}^n\to\mathbb{C}^m$ cannot see this vector. Thus the PBH matrix loses rank at $\lambda$ exactly when $A$ has an eigenvector with eigenvalue $\lambda$ that lies in $\ker C$. [/guided] [/step] [step:Identify invisible eigenvectors with eigenvectors on the unobservable subspace] Let $\lambda \in \mathbb{C}$. Suppose first that the rank condition fails at $\lambda$. By the previous step, there exists $v \in \mathbb{C}^n\setminus\{0\}$ such that $Av=\lambda v$ and $Cv=0$. For every integer $k$ with $0 \le k \le n-1$, \begin{align*} CA^k v = C(\lambda^k v)=\lambda^k Cv=0. \end{align*} Hence $v \in \mathcal{N}$. Since $Bv=Av=\lambda v$, the scalar $\lambda$ is an eigenvalue of $B=A|_{\mathcal{N}}$. Conversely, suppose $\lambda$ is an eigenvalue of $B$. Then there exists $v \in \mathcal{N}\setminus\{0\}$ such that \begin{align*} Bv=\lambda v. \end{align*} Because $Bv=Av$ and $v \in \mathcal{N}\subseteq \ker C$, we have \begin{align*} Av=\lambda v \end{align*} and \begin{align*} Cv=0. \end{align*} Thus $v$ lies in the kernel of $M(\lambda)$, so $\operatorname{rank} M(\lambda)<n$. Let $\sigma(A|_{\mathcal{N}})$ denote the spectrum of the restricted linear operator $A|_{\mathcal{N}}:\mathcal{N}\to\mathcal{N}$, and let $I_{\mathcal{N}}:\mathcal{N}\to\mathcal{N}$ denote the identity map on $\mathcal{N}$. Since $\mathcal{N}$ is a finite-dimensional complex vector space, a scalar $\mu \in \mathbb{C}$ lies in $\sigma(A|_{\mathcal{N}})$ if and only if $A|_{\mathcal{N}}-\mu I_{\mathcal{N}}$ is not invertible, which is equivalent in finite dimension to the existence of a nonzero vector $w \in \mathcal{N}$ satisfying $(A|_{\mathcal{N}})w=\mu w$. Therefore the rank condition fails at $\lambda$ if and only if $\lambda \in \sigma(A|_{\mathcal{N}})$. [/step] [step:Conclude that the right half-plane rank test is equivalent to detectability] By the preceding step, for each $\lambda \in \mathbb{C}$, \begin{align*} \operatorname{rank} M(\lambda)<n \end{align*} if and only if $\lambda \in \sigma(A|_{\mathcal{N}})$. Therefore \begin{align*} \operatorname{rank} M(\lambda)=n \end{align*} for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \ge 0$ if and only if \begin{align*} \sigma(A|_{\mathcal{N}})\cap \{\lambda \in \mathbb{C}:\operatorname{Re}\lambda \ge 0\}=\varnothing. \end{align*} This is equivalent to \begin{align*} \sigma(A|_{\mathcal{N}})\subset \{\lambda \in \mathbb{C}:\operatorname{Re}\lambda < 0\}. \end{align*} By the definition of detectability stated in the theorem, namely that the spectrum of $A|_{\mathcal{N}}$ is contained in the open left half-plane, this proves that $(C,A)$ is detectable if and only if the PBH rank condition holds on the closed right half-plane. [/step]