[proofplan] We work entirely in Kalman coordinates, where the input map has no component in the unreachable subspace. After transforming the feedback law into these coordinates, the resulting feedback perturbation therefore has zero unreachable component, so the unreachable part of the closed-loop operator is unchanged. Finally, block triangularity implies that the characteristic polynomial factors through the diagonal blocks, so the eigenvalues coming from the unreachable diagonal blocks are invariant under feedback. [/proofplan] [step:Transform the feedback law into Kalman coordinates] Let $A_K:\mathbb{R}^n\to\mathbb{R}^n$ denote the closed-loop matrix associated to the feedback law $u=Kx+v$, so $A_K:=A+BK$. In Kalman coordinates, let $R\subset\mathbb{R}^n$ denote the reachable subspace, let $U$ denote the chosen unreachable complement, and let $T:R\oplus U\to\mathbb{R}^n$ denote the invertible coordinate map. Define the transformed feedback map $\tilde K: R \oplus U \to \mathbb{R}^m$ by $\tilde K := KT$. Since $A_K = A + BK$, conjugating by $T$ gives \begin{align*} T^{-1}A_KT = T^{-1}AT + T^{-1}BKT. \end{align*} Using the definitions $\tilde A := T^{-1}AT$, $\tilde B := T^{-1}B$, and $\tilde K := KT$, this becomes \begin{align*} T^{-1}A_KT = \tilde A + \tilde B\tilde K. \end{align*} Thus the closed-loop dynamics in Kalman coordinates are governed by $\tilde A + \tilde B\tilde K$. [guided] Let $A_K:\mathbb{R}^n\to\mathbb{R}^n$ denote the closed-loop matrix associated to the feedback law $u=Kx+v$, so $A_K:=A+BK$. In Kalman coordinates, let $R\subset\mathbb{R}^n$ denote the reachable subspace, let $U$ denote the chosen unreachable complement, and let $T:R\oplus U\to\mathbb{R}^n$ denote the invertible coordinate map. The feedback law is originally written in the original state variable $x \in \mathbb{R}^n$. In Kalman coordinates we write $x = Tz$, where $z \in R \oplus U$. Therefore the feedback term becomes \begin{align*} Kx = K(Tz) = (KT)z. \end{align*} This is why we define $\tilde K := KT$ as a [linear map](/page/Linear%20Map) from $R \oplus U$ to $\mathbb{R}^m$. Now compute the closed-loop operator in the new coordinates. Since $A_K = A + BK$, conjugation by $T$ gives \begin{align*} T^{-1}A_KT = T^{-1}(A + BK)T. \end{align*} Using linearity and associativity of matrix multiplication, \begin{align*} T^{-1}(A + BK)T = T^{-1}AT + T^{-1}BKT. \end{align*} By definition, $T^{-1}AT = \tilde A$, $T^{-1}B = \tilde B$, and $KT = \tilde K$. Hence \begin{align*} T^{-1}A_KT = \tilde A + \tilde B\tilde K. \end{align*} So feedback changes the Kalman-coordinate matrix only through the additive term $\tilde B\tilde K$. [/guided] [/step] [step:Show that the feedback perturbation has no unreachable component] Let $P_U: R \oplus U \to U$ denote the projection $P_U(r,u)=u$. By the defining Kalman-coordinate form, the transformed input map has image contained in the reachable summand $R$. Equivalently, there is a linear map $B_R:\mathbb{R}^m\to R$ such that $\tilde B:R\oplus U\leftarrow \mathbb{R}^m$ is given by $\tilde B w=(B_Rw,0)$ for every $w\in\mathbb{R}^m$. For every $z \in R \oplus U$, the vector $\tilde Kz$ lies in $\mathbb{R}^m$, so this Kalman-coordinate property gives \begin{align*} \tilde B\tilde Kz = (B_R\tilde Kz,0). \end{align*} Therefore \begin{align*} P_U(\tilde B\tilde Kz)=0. \end{align*} Thus the perturbation $\tilde B\tilde K$ has zero component in the unreachable subspace $U$. [/step] [step:Identify the unchanged unreachable block of the closed-loop operator] By the block triangular Kalman-coordinate form of $\tilde A$, there are linear maps $A_R:R\to R$, $C:U\to R$, and $A_U:U\to U$ such that \begin{align*} \tilde A(r,u)=(A_Rr+Cu,A_Uu) \end{align*} for every $(r,u)\in R\oplus U$. For $z=(r,u)\in R\oplus U$, the closed-loop operator in Kalman coordinates therefore satisfies \begin{align*} (\tilde A+\tilde B\tilde K)(r,u)=(A_Rr+Cu+B_R\tilde K(r,u), A_Uu). \end{align*} Hence the $U$-component of $(\tilde A+\tilde B\tilde K)(r,u)$ is exactly $A_Uu$, independent of $K$. Therefore the induced lower unreachable block of $T^{-1}A_KT$ is the same map $A_U:U\to U$ as for $\tilde A$. The phrase "unreachable diagonal blocks" in the statement refers to the diagonal blocks obtained after choosing a block triangular decomposition of the unreachable operator $A_U$. Thus, if $U=U_1\oplus\cdots\oplus U_s$ is the chosen block triangular decomposition for $A_U$, define $A_{U_j}:U_j\to U_j$ for $1\leq j\leq s$ to be the diagonal block induced by $A_U$ on $U_j$. Since the whole lower unreachable operator remains $A_U$, each diagonal map on this chosen unreachable block decomposition is unchanged. Hence the unreachable diagonal blocks of $T^{-1}A_KT$ are precisely $A_{U_1},\dots,A_{U_s}$. [/step] [step:Factor the characteristic polynomial through the block triangular diagonal] Let $M: V_1\oplus\cdots\oplus V_q \to V_1\oplus\cdots\oplus V_q$ be any block triangular linear map on a finite-dimensional real [vector space](/page/Vector%20Space), with finite-dimensional diagonal blocks $M_j:V_j\to V_j$. In the application below, these spaces are finite-dimensional because each $V_j$ is a subspace or direct summand of $R\oplus U$, and $R\oplus U$ is isomorphic to $\mathbb{R}^n$ through $T$. Choose a basis of each $V_j$ and concatenate these bases in the order $V_1,\dots,V_q$; this adapted basis represents $M$ by a finite block triangular matrix with diagonal matrix blocks representing the maps $M_j$. For each $\lambda\in\mathbb{C}$, the map $\lambda I-M$ is therefore represented in the same adapted basis by a finite block triangular matrix with diagonal blocks $\lambda I_{V_j}-M_j$. Expanding the [determinant](/page/Determinant) by the Leibniz formula, every nonzero permutation term of a block triangular matrix preserves the initial block filtration, so it decomposes into permutation terms inside the diagonal blocks. Hence the finite [block triangular determinant](/theorems/399) product formula gives \begin{align*} \det(\lambda I-M)=\prod_{j=1}^{q}\det(\lambda I_{V_j}-M_j). \end{align*} [guided] The point of this step is to isolate which factors of the characteristic polynomial can see the unreachable dynamics. Let $M: V_1\oplus\cdots\oplus V_q \to V_1\oplus\cdots\oplus V_q$ be a block triangular linear map on a finite-dimensional real vector space, and let $M_j:V_j\to V_j$ denote its diagonal block on $V_j$. Each $V_j$ is finite-dimensional in the application because $V_j$ is a subspace or direct summand of $R\oplus U$, and $R\oplus U$ is isomorphic to $\mathbb{R}^n$ by the invertible coordinate map $T$. Choose a basis of each $V_j$ and concatenate these bases in the order $V_1,\dots,V_q$. In this adapted basis, the matrix of $M$ is block triangular, with the matrix of $M_j$ on the $j$th diagonal block. For each $\lambda\in\mathbb{C}$, the matrix of $\lambda I-M$ in the same adapted basis is again block triangular, and its $j$th diagonal block is $\lambda I_{V_j}-M_j$. Expanding the [determinant](/page/Determinant) by the Leibniz formula, every nonzero permutation term of a block triangular matrix preserves the initial block filtration, so it decomposes into permutation terms inside the diagonal blocks. Therefore the finite block triangular determinant product formula gives \begin{align*} \det(\lambda I-M)=\prod_{j=1}^{q}\det(\lambda I_{V_j}-M_j). \end{align*} This formula is exactly what is needed: the characteristic polynomial of a block triangular operator factors as the product of the characteristic polynomials of its diagonal blocks. Now apply this factorization to the closed-loop Kalman matrix $T^{-1}A_KT$. Its full block triangular diagonal consists of the reachable diagonal block followed by the unchanged unreachable diagonal blocks $A_{U_1},\dots,A_{U_s}$. Hence the factors in the characteristic polynomial coming from the unreachable diagonal blocks are exactly \begin{align*} \prod_{j=1}^{s}\det(\lambda I_{U_j}-A_{U_j}). \end{align*} These factors are the same as for $T^{-1}AT$, because the previous step proved that the entire lower unreachable operator remains $A_U$ after feedback. It remains to relate the Kalman-coordinate matrices to the original matrices. Since $T:R\oplus U\to\mathbb{R}^n$ is an invertible linear map between finite-dimensional real vector spaces, the matrices $A_K$ and $T^{-1}A_KT$ are similar, and the matrices $A$ and $T^{-1}AT$ are similar. We use the multiplicativity of the [determinant](/page/Determinant) to verify directly that characteristic polynomials are unchanged under this similarity: \begin{align*} \det(\lambda I-T^{-1}MT)=\det(T^{-1}(\lambda I-M)T)=\det(T^{-1})\det(\lambda I-M)\det(T)=\det(\lambda I-M) \end{align*} for every finite-dimensional matrix $M$ of compatible size. Thus $A_K$ and $T^{-1}A_KT$ have the same eigenvalues, and $A$ and $T^{-1}AT$ have the same eigenvalues. Therefore the eigenvalues contributed by the unreachable diagonal blocks are unchanged by the feedback matrix $K$. [/guided] In the closed-loop Kalman matrix $T^{-1}A_KT$, the full block triangular diagonal consists of the reachable diagonal block followed by the unchanged unreachable diagonal blocks $A_{U_1},\dots,A_{U_s}$. Hence the factors coming from the unreachable diagonal blocks are exactly \begin{align*} \prod_{j=1}^{s}\det(\lambda I_{U_j}-A_{U_j}). \end{align*} These factors are the same as for $T^{-1}AT$. Since $T:R\oplus U\to\mathbb{R}^n$ is an invertible linear map between finite-dimensional real vector spaces, the matrices $A_K$ and $T^{-1}A_KT$ are similar, and the matrices $A$ and $T^{-1}AT$ are similar. The characteristic polynomial is invariant under similarity because \begin{align*} \det(\lambda I-T^{-1}MT)=\det(T^{-1}(\lambda I-M)T)=\det(T^{-1})\det(\lambda I-M)\det(T)=\det(\lambda I-M) \end{align*} for every finite-dimensional matrix $M$ of compatible size, using multiplicativity of the [determinant](/page/Determinant). Thus $A_K$ and $T^{-1}A_KT$ have the same eigenvalues, and $A$ and $T^{-1}AT$ have the same eigenvalues. Therefore the eigenvalues contributed by the unreachable diagonal blocks are unchanged by the feedback matrix $K$. [/step]