[proofplan] We use the finite-dimensional rank tests for reachability and observability, namely the Kalman controllability and observability rank criteria, then compare the associated reachability and observability matrices directly. The similarity identities imply $\tilde A^k\tilde B=T^{-1}A^kB$ and $\tilde C\tilde A^k=CA^kT$ for every relevant integer $k$. Therefore the transformed reachability matrix is obtained by left multiplication by the invertible matrix $T^{-1}$, while the transformed observability matrix is obtained by right multiplication by the invertible matrix $T$. Since invertible multiplication preserves rank, the rank tests for reachability and observability are unchanged. [/proofplan]

[step:Compute how the reachable columns transform under similarity]Let $n$ denote the state dimension, let $m$ denote the input dimension, and let $p$ denote the output dimension, so $A,\tilde A,T \in \mathbb{R}^{n \times n}$ with $T$ invertible, $B,\tilde B \in \mathbb{R}^{n \times m}$, and $C,\tilde C \in \mathbb{R}^{p \times n}$. Let $I_n \in \mathbb{R}^{n \times n}$ denote the identity matrix. For each integer $k$ with $0 \le k \le n-1$, we claim \begin{align*} \tilde A^k\tilde B = T^{-1}A^kB. \end{align*} For $k=0$, this is precisely $\tilde B=T^{-1}B$, using the convention $A^0=I_n$ and $\tilde A^0=I_n$. For $k \ge 1$, the identity follows by cancellation of adjacent factors $TT^{-1}$: \begin{align*} \tilde A^k\tilde B = (T^{-1}AT)^kT^{-1}B = T^{-1}A^kB. \end{align*} Define the reachability matrices $\mathcal R(A,B) \in \mathbb{R}^{n \times nm}$ and $\mathcal R(\tilde A,\tilde B) \in \mathbb{R}^{n \times nm}$ by horizontal block concatenation: \begin{align*} \mathcal R(A,B) := [B, AB, \ldots, A^{n-1}B]. \end{align*} Also define \begin{align*} \mathcal R(\tilde A,\tilde B) := [\tilde B, \tilde A\tilde B, \ldots, \tilde A^{n-1}\tilde B]. \end{align*} Hence, block column by block column, \begin{align*} \mathcal R(\tilde A,\tilde B) = T^{-1}\mathcal R(A,B). \end{align*}[/step]

[guided]Let $n$ denote the state dimension, let $m$ denote the input dimension, and let $p$ denote the output dimension, so $A,\tilde A,T \in \mathbb{R}^{n \times n}$ with $T$ invertible, $B,\tilde B \in \mathbb{R}^{n \times m}$, and $C,\tilde C \in \mathbb{R}^{p \times n}$. The reachability matrix is built from the columns of $B,AB,\dots,A^{n-1}B$. Thus we need to understand how each block $A^kB$ changes when the state coordinate is replaced by the similar coordinate system determined by $T$. For $k=0$, the transformed block is just \begin{align*} \tilde A^0\tilde B = I_n\tilde B = \tilde B = T^{-1}B = T^{-1}A^0B. \end{align*} Now let $k$ be an integer with $1 \le k \le n-1$. Since $\tilde A=T^{-1}AT$, multiplying powers of $\tilde A$ produces adjacent factors $TT^{-1}$, each of which equals $I_n$. Therefore \begin{align*} \tilde A^k\tilde B = (T^{-1}AT)^kT^{-1}B. \end{align*} Cancelling the adjacent inverse pairs gives \begin{align*} (T^{-1}AT)^kT^{-1}B = T^{-1}A^kB. \end{align*} So every reachable block in the transformed realization is obtained from the corresponding original block by the same left multiplication by $T^{-1}$. Define the reachability matrices $\mathcal R(A,B) \in \mathbb{R}^{n \times nm}$ and $\mathcal R(\tilde A,\tilde B) \in \mathbb{R}^{n \times nm}$ by \begin{align*} \mathcal R(A,B) := [B, AB, \ldots, A^{n-1}B] \end{align*} and \begin{align*} \mathcal R(\tilde A,\tilde B) := [\tilde B, \tilde A\tilde B, \ldots, \tilde A^{n-1}\tilde B]. \end{align*} Since the reachability matrix is the horizontal concatenation of these blocks, this gives \begin{align*} \mathcal R(\tilde A,\tilde B) = T^{-1}\mathcal R(A,B). \end{align*}[/guided]

[step:Use invertibility of $T$ to preserve reachability rank]Let $\operatorname{rank} M$ denote the column rank of a real matrix $M$. Since $T^{-1}$ is invertible, the [linear map](/page/Linear%20Map) $x \mapsto T^{-1}x$ is a bijection from $\mathbb{R}^n$ to $\mathbb{R}^n$. Therefore it maps the column space of $\mathcal R(A,B)$ bijectively onto the column space of $T^{-1}\mathcal R(A,B)$. Hence \begin{align*} \operatorname{rank}\mathcal R(\tilde A,\tilde B) = \operatorname{rank}\mathcal R(A,B). \end{align*} By the Kalman controllability rank criterion, $(A,B)$ is reachable exactly when $\operatorname{rank}\mathcal R(A,B)=n$, and $(\tilde A,\tilde B)$ is reachable exactly when $\operatorname{rank}\mathcal R(\tilde A,\tilde B)=n$. The matrices are real finite-dimensional matrices of the required sizes, so the test applies. The equality of ranks gives the equivalence.[/step]

[guided]We have already proved that the transformed reachability matrix satisfies \begin{align*} \mathcal R(\tilde A,\tilde B)=T^{-1}\mathcal R(A,B). \end{align*} The remaining question is whether left multiplication by $T^{-1}$ can change the number of independent reachable columns. It cannot, because $T^{-1}:\mathbb R^n\to\mathbb R^n$ is an invertible linear map. Therefore it sends a set of columns in $\mathbb R^n$ to another set with exactly the same linear-dependence relations: a linear combination of the columns of $\mathcal R(A,B)$ is zero exactly when the corresponding linear combination after applying $T^{-1}$ is zero. Equivalently, it maps the column space of $\mathcal R(A,B)$ bijectively onto the column space of $T^{-1}\mathcal R(A,B)$. Thus the column-space dimensions agree: \begin{align*} \operatorname{rank}\mathcal R(\tilde A,\tilde B)=\operatorname{rank}(T^{-1}\mathcal R(A,B))=\operatorname{rank}\mathcal R(A,B). \end{align*} The Kalman controllability rank criterion applies because both reachability matrices are real finite-dimensional matrices with $n$ rows. It says that $(A,B)$ is reachable exactly when \begin{align*} \operatorname{rank}\mathcal R(A,B)=n, \end{align*} and that $(\tilde A,\tilde B)$ is reachable exactly when \begin{align*} \operatorname{rank}\mathcal R(\tilde A,\tilde B)=n. \end{align*} Since the two ranks are equal, one condition holds exactly when the other holds. Hence $(A,B)$ is reachable if and only if $(\tilde A,\tilde B)$ is reachable.[/guided]

[step:Compute how the observable rows transform under similarity]For each integer $k$ with $0 \le k \le n-1$, we claim \begin{align*} \tilde C\tilde A^k = CA^kT. \end{align*} For $k=0$, this is $\tilde C=CT=CA^0T$. For $k \ge 1$, \begin{align*} \tilde C\tilde A^k = CT(T^{-1}AT)^k = CA^kT. \end{align*} Define the observability matrices $\mathcal O(C,A) \in \mathbb{R}^{pn \times n}$ and $\mathcal O(\tilde C,\tilde A) \in \mathbb{R}^{pn \times n}$ as vertical block concatenations by \begin{align*} \mathcal O(C,A) := \operatorname{col}(C, CA, \ldots, CA^{n-1}) \end{align*} and \begin{align*} \mathcal O(\tilde C,\tilde A) := \operatorname{col}(\tilde C, \tilde C\tilde A, \ldots, \tilde C\tilde A^{n-1}), \end{align*} where $\operatorname{col}$ denotes vertical block concatenation. Thus the vertical block concatenation defining the observability matrix satisfies \begin{align*} \mathcal O(\tilde C,\tilde A) = \mathcal O(C,A)T. \end{align*}[/step]

[guided]The observability matrix is built from the output rows $C,CA,\ldots,CA^{n-1}$. We therefore compute how each block row changes under the similarity transformation. For $k=0$, the identity follows directly from the similarity relation for the output matrix: \begin{align*} \tilde C\tilde A^0=\tilde C=CT=CA^0T. \end{align*} Now let $k$ be an integer with $1\le k\le n-1$. Using $\tilde C=CT$ and $\tilde A=T^{-1}AT$, we obtain \begin{align*} \tilde C\tilde A^k=CT(T^{-1}AT)^k. \end{align*} The adjacent factors $TT^{-1}$ cancel in the product, so this becomes \begin{align*} CT(T^{-1}AT)^k=CA^kT. \end{align*} Thus each transformed observable block row is obtained from the corresponding original block row by the same right multiplication by $T$. Define the observability matrices $\mathcal O(C,A)\in\mathbb R^{pn\times n}$ and $\mathcal O(\tilde C,\tilde A)\in\mathbb R^{pn\times n}$ by vertical block concatenation: \begin{align*} \mathcal O(C,A):=\operatorname{col}(C,CA,\ldots,CA^{n-1}) \end{align*} and \begin{align*} \mathcal O(\tilde C,\tilde A):=\operatorname{col}(\tilde C,\tilde C\tilde A,\ldots,\tilde C\tilde A^{n-1}), \end{align*} where $\operatorname{col}$ denotes vertical block concatenation. Since every block row has acquired the same right factor $T$, the whole observability matrix satisfies \begin{align*} \mathcal O(\tilde C,\tilde A)=\mathcal O(C,A)T. \end{align*}[/guided]

[step:Use invertibility of $T$ to preserve observability rank]Since $T$ is invertible, right multiplication by $T$ is an invertible column-coordinate change on matrices with $n$ columns. For any matrix $M$ with $n$ columns, define its null space by \begin{align*} \ker M:=\{x\in\mathbb R^n: Mx=0\}. \end{align*} Then the null spaces of $M$ and $MT$ have the same dimension because \begin{align*} MTx=0 \iff M(Tx)=0. \end{align*} The map $x \mapsto Tx$ is a bijection on $\mathbb{R}^n$, so $\dim\ker(MT)=\dim\ker M$, and the [Rank-Nullity Theorem](/theorems/385) gives $\operatorname{rank}(MT)=\operatorname{rank}M$. Applying this to $M=\mathcal O(C,A)$ yields \begin{align*} \operatorname{rank}\mathcal O(\tilde C,\tilde A) = \operatorname{rank}\mathcal O(C,A). \end{align*} By the Kalman observability rank criterion, $(C,A)$ is observable exactly when $\operatorname{rank}\mathcal O(C,A)=n$, and $(\tilde C,\tilde A)$ is observable exactly when $\operatorname{rank}\mathcal O(\tilde C,\tilde A)=n$. The matrices are real finite-dimensional matrices of the required sizes, so the test applies. The equality of ranks gives the desired equivalence, completing the proof.[/step]

[guided]We have proved that the transformed observability matrix is \begin{align*} \mathcal O(\tilde C,\tilde A)=\mathcal O(C,A)T. \end{align*} We now show that multiplying on the right by the invertible matrix $T$ does not change rank. Let $M$ be any real matrix with $n$ columns, and define its null space by \begin{align*} \ker M:=\{x\in\mathbb R^n: Mx=0\}. \end{align*} For $x\in\mathbb R^n$, the equation $(MT)x=0$ is exactly the equation $M(Tx)=0$: \begin{align*} MTx=0\iff M(Tx)=0. \end{align*} Because $T:\mathbb R^n\to\mathbb R^n$ is bijective, the map $x\mapsto Tx$ gives a bijection from $\ker(MT)$ onto $\ker M$. Therefore \begin{align*} \dim\ker(MT)=\dim\ker M. \end{align*} The [Rank-Nullity Theorem](/theorems/385) for real finite-dimensional matrices gives \begin{align*} \operatorname{rank}(MT)+\dim\ker(MT)=n \end{align*} and \begin{align*} \operatorname{rank}M+\dim\ker M=n. \end{align*} Since the nullities agree, the ranks agree: \begin{align*} \operatorname{rank}(MT)=\operatorname{rank}M. \end{align*} Applying this with $M=\mathcal O(C,A)$ yields \begin{align*} \operatorname{rank}\mathcal O(\tilde C,\tilde A)=\operatorname{rank}\mathcal O(C,A). \end{align*} The Kalman observability rank criterion applies because both observability matrices are real finite-dimensional matrices with $n$ columns. It says that $(C,A)$ is observable exactly when \begin{align*} \operatorname{rank}\mathcal O(C,A)=n, \end{align*} and that $(\tilde C,\tilde A)$ is observable exactly when \begin{align*} \operatorname{rank}\mathcal O(\tilde C,\tilde A)=n. \end{align*} The ranks are equal, so one condition holds exactly when the other holds. Hence $(C,A)$ is observable if and only if $(\tilde C,\tilde A)$ is observable, completing the proof.[/guided]