[proofplan] We prove equality of the two subspaces by proving both inclusions through their finite-dimensional structure. The inclusion $\mathcal{R}_T \subseteq \operatorname{Range}[B\ AB\ \cdots\ A^{n-1}B]$ follows from the [Cayley-Hamilton theorem](/theorems/865), which expresses every matrix exponential $e^{At}$ as a linear combination of $I,A,\dots,A^{n-1}$. For the reverse inclusion, we compare orthogonal complements: a vector orthogonal to all reachable states is orthogonal to $e^{A(T-s)}B$ for every time $s$, and analyticity then lets us differentiate at $s=T$ to obtain orthogonality to $B,AB,\dots,A^{n-1}B$. [/proofplan] [step:Define the controllability matrix and the target subspace] Define the controllability matrix $\mathcal{C} \in \mathbb{R}^{n \times nm}$ by \begin{align*} \mathcal{C} := \begin{bmatrix} B & AB & A^2B & \cdots & A^{n-1}B \end{bmatrix}. \end{align*} Let $V \subseteq \mathbb{R}^n$ denote its range: \begin{align*} V := \operatorname{Range}\mathcal{C}. \end{align*} Equivalently, $V$ is the linear span of all vectors $A^kBv$ with $k \in \{0,\dots,n-1\}$ and $v \in \mathbb{R}^m$. We prove $\mathcal{R}_T = V$ for an arbitrary fixed $T > 0$. [/step] [step:Use Cayley-Hamilton to put every reachable vector in the controllability range] For each $t \in [0,T]$, the [Cayley-Hamilton theorem](/theorems/923) gives scalar functions $a_0,\dots,a_{n-1}: [0,T] \to \mathbb{R}$ such that \begin{align*} e^{At} = \sum_{k=0}^{n-1} a_k(t) A^k. \end{align*} Here we use the standard consequence of Cayley-Hamilton that the polynomial functional calculus modulo the characteristic polynomial reduces every [power series](/page/Power%20Series) in $A$ to a polynomial in $A$ of degree at most $n-1$. Let $u \in L^1((0,T);\mathbb{R}^m)$, and define $x_u \in \mathbb{R}^n$ by \begin{align*} x_u := \int_0^{\!T}e^{A(T-s)}B u(s)\, d\mathcal{L}^1(s). \end{align*} Substituting the Cayley-Hamilton representation with $t=T-s$ gives \begin{align*} x_u = \sum_{k=0}^{n-1} A^kB \left(\int_0^{\!T}a_k(T-s)u(s)\, d\mathcal{L}^1(s)\right). \end{align*} The vector inside the parentheses belongs to $\mathbb{R}^m$, so every summand belongs to $\operatorname{Range}(A^kB) \subseteq V$. Since $V$ is a linear subspace, $x_u \in V$. Therefore $\mathcal{R}_T \subseteq V$. [guided] We first show that no reachable vector can point outside the span of the columns of $B,AB,\dots,A^{n-1}B$. The reason is that the matrix exponential cannot create genuinely new powers of $A$ beyond degree $n-1$: by the Cayley-Hamilton theorem, every sufficiently high power of $A$ is a linear combination of $I,A,\dots,A^{n-1}$. Applying this reduction to the power series for $e^{At}$ gives scalar functions $a_0,\dots,a_{n-1}: [0,T] \to \mathbb{R}$ satisfying \begin{align*} e^{At} = \sum_{k=0}^{n-1} a_k(t) A^k \end{align*} for every $t \in [0,T]$. Now take an arbitrary control $u \in L^1((0,T);\mathbb{R}^m)$ and define its terminal state $x_u \in \mathbb{R}^n$ by \begin{align*} x_u := \int_0^{\!T}e^{A(T-s)}B u(s)\, d\mathcal{L}^1(s). \end{align*} The integrand is integrable because $s \mapsto e^{A(T-s)}B$ is continuous on the compact interval $[0,T]$ and hence bounded, while $u \in L^1((0,T);\mathbb{R}^m)$. Substituting the representation of $e^{A(T-s)}$ gives \begin{align*} x_u = \int_0^{\!T}\sum_{k=0}^{n-1} a_k(T-s)A^kB u(s)\, d\mathcal{L}^1(s). \end{align*} Linearity of the finite-dimensional Bochner integral allows the finite sum and the fixed matrices $A^kB$ to be pulled outside the integral, so \begin{align*} x_u = \sum_{k=0}^{n-1} A^kB \left(\int_0^{\!T}a_k(T-s)u(s)\, d\mathcal{L}^1(s)\right). \end{align*} For each $k$, the integral in parentheses is a vector in $\mathbb{R}^m$. Hence $A^kB$ applied to that vector lies in $\operatorname{Range}(A^kB)$, which is contained in $V$. Therefore $x_u$ is a finite sum of vectors in $V$, and since $V$ is a subspace, $x_u \in V$. Because $u$ was arbitrary, $\mathcal{R}_T \subseteq V$. [/guided] [/step] [step:Translate orthogonality to reachability into pointwise orthogonality of the kernel] We prove the reverse inclusion by comparing orthogonal complements in the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^n$. Let $q \in \mathbb{R}^n$ satisfy $q \perp \mathcal{R}_T$, meaning \begin{align*} q^\top x = 0 \end{align*} for every $x \in \mathcal{R}_T$. Define the continuous matrix-valued function $G: [0,T] \to \mathbb{R}^{1 \times m}$ by \begin{align*} G(s) := q^\top e^{A(T-s)}B. \end{align*} For every $u \in L^1((0,T);\mathbb{R}^m)$, \begin{align*} 0 = q^\top \int_0^{\!T}e^{A(T-s)}B u(s)\, d\mathcal{L}^1(s) = \int_0^{\!T}G(s)u(s)\, d\mathcal{L}^1(s). \end{align*} We claim that $G(s)=0$ for every $s \in [0,T]$. If not, there exist $s_0 \in [0,T]$, an index $j \in \{1,\dots,m\}$, and $\varepsilon > 0$ such that the $j$th component $G_j$ has constant nonzero sign on a measurable interval $I \subset (0,T)$ with positive $\mathcal{L}^1$-measure and $|G_j(s)| \ge \varepsilon$ for all $s \in I$. Let $e_j \in \mathbb{R}^m$ be the $j$th standard basis vector and define $u_I: (0,T) \to \mathbb{R}^m$ by \begin{align*} u_I(s) := \operatorname{sgn}(G_j(s))\mathbb{1}_I(s)e_j. \end{align*} Then $u_I \in L^1((0,T);\mathbb{R}^m)$ and \begin{align*} 0 = \int_0^{\!T}G(s)u_I(s)\, d\mathcal{L}^1(s) = \int_I |G_j(s)|\, d\mathcal{L}^1(s) \ge \varepsilon \mathcal{L}^1(I) > 0, \end{align*} a contradiction. Thus $G(s)=0$ for all $s \in [0,T]$. [/step] [step:Differentiate at the terminal time to annihilate the columns of the controllability matrix] Since $s \mapsto q^\top e^{A(T-s)}B$ is identically zero on $[0,T]$, all of its derivatives vanish at $s=T$. For each integer $k \ge 0$, differentiating $k$ times gives \begin{align*} \frac{d^k}{ds^k}\left(q^\top e^{A(T-s)}B\right)\bigg|_{s=T} = (-1)^k q^\top A^kB = 0. \end{align*} In particular, \begin{align*} q^\top A^kB = 0 \end{align*} for every $k \in \{0,\dots,n-1\}$. Hence $q$ is orthogonal to every column of $\mathcal{C}$, so $q \in V^\perp$. We have proved $\mathcal{R}_T^\perp \subseteq V^\perp$. [/step] [step:Conclude equality from equality of orthogonal complements] The inclusion $\mathcal{R}_T \subseteq V$ proved earlier implies $V^\perp \subseteq \mathcal{R}_T^\perp$. The previous step gives the reverse inclusion $\mathcal{R}_T^\perp \subseteq V^\perp$, hence \begin{align*} \mathcal{R}_T^\perp = V^\perp. \end{align*} Both $\mathcal{R}_T$ and $V$ are linear subspaces of the finite-dimensional Euclidean space $\mathbb{R}^n$, so taking orthogonal complements twice gives \begin{align*} \mathcal{R}_T = \left(\mathcal{R}_T^\perp\right)^\perp = \left(V^\perp\right)^\perp = V. \end{align*} Substituting the definition of $V$ yields \begin{align*} \mathcal{R}_T = \operatorname{Range}\begin{bmatrix} B & AB & A^2B & \cdots & A^{n-1}B \end{bmatrix}. \end{align*} This proves the formula for the fixed $T>0$, and since $T$ was arbitrary, the formula holds for every $T>0$. [/step]