[proofplan] We work entirely in Kalman coordinates. In those coordinates, the output matrix $\tilde C$ has zero columns on the unobservable coordinates, so the product $\tilde L\tilde C$ has a zero unobservable-to-observable column block and a zero unobservable diagonal block. Subtracting this product from $\tilde A$ therefore changes only the observable-column dynamics, leaving the unobservable diagonal block unchanged. The spectral statement follows by factoring the characteristic polynomial of the resulting block triangular matrix. [/proofplan] [step:Transform the observer error matrix into Kalman coordinates] Since $T \in \mathbb{R}^{n \times n}$ is the invertible Kalman-coordinate change matrix, conjugating the observer error matrix $A-LC$ by $T$ gives $T^{-1}(A-LC)T = T^{-1}AT - T^{-1}LCT$. By the definitions $\tilde A := T^{-1}AT$, $\tilde L := T^{-1}L$, and $\tilde C := CT$, this becomes $T^{-1}(A-LC)T = \tilde A - \tilde L\tilde C$. Thus it is enough to compute the block form of $\tilde A-\tilde L\tilde C$. [/step] [step:Compute the injection term and locate its zero unobservable block] Using the block decompositions, write $\tilde L$ as the block column with first block $L_o$ and second block $L_u$, and write $\tilde C$ as the block row whose first block is $C_o$ and whose second block is the zero matrix. This avoids introducing any additional row notation for a one-row block matrix. Block matrix multiplication gives the four block identities. The upper-left block is $(\tilde L\tilde C)_{11}=L_oC_o$. The upper-right block is $(\tilde L\tilde C)_{12}=0$. The lower-left block is $(\tilde L\tilde C)_{21}=L_uC_o$. The lower-right block is $(\tilde L\tilde C)_{22}=0$. The zero block $(\tilde L\tilde C)_{12}$ has size $r \times (n-r)$, and the zero block $(\tilde L\tilde C)_{22}$ has size $(n-r) \times (n-r)$. Therefore the observer injection term has no component acting from the unobservable coordinates into either diagonal unobservable dynamics. [guided] The point of Kalman observability coordinates is that the output matrix ignores the unobservable coordinates. Algebraically, this means that $\tilde C$ is the block row whose first block is $C_o$ and whose second block is the zero matrix. Now take an arbitrary transformed gain $\tilde L$, written as the block column with first block $L_o$ and second block $L_u$. Here $L_o$ maps output errors into the observable coordinate equations, while $L_u$ maps output errors into the unobservable coordinate equations. Multiplying the two block matrices gives four identities: $(\tilde L\tilde C)_{11}=L_oC_o$, $(\tilde L\tilde C)_{12}=0$, $(\tilde L\tilde C)_{21}=L_uC_o$, and $(\tilde L\tilde C)_{22}=0$. The decisive feature is the zero second block column. No choice of $L_o$ or $L_u$ can create a nonzero term in that second block column, because the second block column of $\tilde C$ is already zero. In particular, the lower-right block of $\tilde L\tilde C$ is the zero matrix in $\mathbb{R}^{(n-r) \times (n-r)}$, so subtracting $\tilde L\tilde C$ cannot alter the unobservable diagonal block. [/guided] [/step] [step:Subtract the injection term from the Kalman form of $A$] By hypothesis, $\tilde A$ has block entries $\tilde A_{11}=A_o$, $\tilde A_{12}=0$, $\tilde A_{21}=A_{uo}$, and $\tilde A_{22}=A_u$. Combining this with the previous computation yields the block identities $(\tilde A-\tilde L\tilde C)_{11}=A_o-L_oC_o$, $(\tilde A-\tilde L\tilde C)_{12}=0$, $(\tilde A-\tilde L\tilde C)_{21}=A_{uo}-L_uC_o$, and $(\tilde A-\tilde L\tilde C)_{22}=A_u$. Since $T^{-1}(A-LC)T = \tilde A-\tilde L\tilde C$, this proves the asserted block form of the transformed observer error matrix. [/step] [step:Factor the characteristic polynomial through the diagonal blocks] Let $\lambda \in \mathbb{C}$ be the spectral parameter, and define the complex block matrix $M_L(\lambda) := \lambda I_n - T^{-1}(A-LC)T$. Using the block form already proved, $M_L(\lambda)$ has block entries $(M_L(\lambda))_{11}=\lambda I_r - (A_o-L_oC_o)$, $(M_L(\lambda))_{12}=0$, $(M_L(\lambda))_{21}=-(A_{uo}-L_uC_o)$, and $(M_L(\lambda))_{22}=\lambda I_{n-r}-A_u$. The [determinant formula for block triangular matrices](/page/Determinant) applies because the upper-right block of $M_L(\lambda)$ is zero and the diagonal blocks are square matrices of sizes $r \times r$ and $(n-r) \times (n-r)$, respectively. In the degenerate cases $r=0$ or $r=n$, this is the same determinant identity with the empty diagonal block contributing the factor $1$. Hence its determinant is the product of the determinants of its diagonal blocks, namely $\det M_L(\lambda)=\det(\lambda I_r - (A_o-L_oC_o))\det(\lambda I_{n-r}-A_u)$. Thus the characteristic polynomial of $T^{-1}(A-LC)T$ contains the unobservable factor $\det(\lambda I_{n-r}-A_u)$, which does not involve $L$. [/step] [step:Conclude that observer injection leaves the unobservable eigenvalues unchanged] Because $T$ is invertible, the characteristic polynomial is unchanged under this similarity: for every $\lambda \in \mathbb{C}$, [determinant multiplicativity](/theorems/395) gives $\det(\lambda I_n-T^{-1}(A-LC)T)=\det(T^{-1}(\lambda I_n-(A-LC))T)=\det(\lambda I_n-(A-LC))$. Therefore $A-LC$ and $T^{-1}(A-LC)T$ have the same eigenvalues, counted with algebraic multiplicity. The preceding factorization shows that the factor $\det(\lambda I_{n-r}-A_u)$ contributed by the unobservable diagonal block is unchanged by $L$. If an eigenvalue also occurs in the observable diagonal block, then its algebraic multiplicity in the full matrix is the sum of the contributions from the diagonal factors; the assertion here is the block-level statement that the unobservable contribution is unchanged. Suppose, more generally, that the unobservable block $A_u$ is written in block triangular form with diagonal blocks $B_1,\dots,B_m$, where each $B_j \in \mathbb{R}^{n_j \times n_j}$ and $n_1+\cdots+n_m=n-r$. Since the lower-right block of $\tilde A-\tilde L\tilde C$ remains the same matrix $A_u$, its internal diagonal blocks remain $B_1,\dots,B_m$. Therefore each unobservable diagonal block has exactly the same eigenvalues before and after observer injection. This proves that observer injection acts only on the observable dynamics in Kalman coordinates. [guided] In Kalman coordinates, the transformed matrices are $\tilde A=T^{-1}AT$, $\tilde L=T^{-1}L$, and $\tilde C=CT$. The observer error matrix transforms by similarity as $T^{-1}(A-LC)T=\tilde A-\tilde L\tilde C$. Since $T$ is invertible, the transformed matrix has the same characteristic polynomial as $A-LC$. Indeed, for every $\lambda \in \mathbb{C}$, determinant multiplicativity gives $\det(\lambda I_n-T^{-1}(A-LC)T)=\det(T^{-1}(\lambda I_n-(A-LC))T)=\det(\lambda I_n-(A-LC))$. The key structural point is that $\tilde C$ has zero columns on the unobservable coordinates. Therefore the second block column of $\tilde L\tilde C$ is zero, regardless of the choice of compatible observer gain $L$. Subtracting $\tilde L\tilde C$ from $\tilde A$ can change the observable block $A_o$ and the lower-left coupling block $A_{uo}$, but it leaves the lower-right unobservable block equal to $A_u$. For the spectral conclusion, define $M_L(\lambda)=\lambda I_n-T^{-1}(A-LC)T$ for $\lambda\in\mathbb C$. Its upper-right block is zero, and its diagonal blocks are square matrices of sizes $r \times r$ and $(n-r) \times (n-r)$, with the empty-block convention giving determinant factor $1$ when $r=0$ or $r=n$. Therefore the determinant formula for block triangular matrices gives $\det M_L(\lambda)=\det(\lambda I_r-(A_o-L_oC_o))\det(\lambda I_{n-r}-A_u)$. The second factor is independent of $L$. Thus the eigenvalue contribution coming from the unobservable block $A_u$ is unchanged by observer injection. If an eigenvalue also appears in the observable factor, its full-matrix algebraic multiplicity includes both contributions, but the contribution from $A_u$ remains fixed. If $A_u$ is itself block triangular with diagonal blocks $B_1,\dots,B_m$, where each $B_j \in \mathbb{R}^{n_j \times n_j}$ and $n_1+\cdots+n_m=n-r$, then those same diagonal blocks remain present after injection, so each unobservable diagonal block has the same eigenvalues before and after the observer gain is applied. The observer gain may still change the lower-left coupling block from $A_{uo}$ to $A_{uo}-L_uC_o$; the invariant statement is precisely about the unobservable diagonal blocks. [/guided] [/step]