[proofplan] We identify the nullspace of the Gramian with the orthogonal complement of the reachable subspace. The key computation rewrites the Gramian quadratic form as the integral of a squared norm, which converts membership in the kernel into a pointwise annihilation condition on the input directions. That condition is then compared directly with the terminal-state formula defining $\mathcal R_T$. Finally, finite-dimensional linear algebra turns the kernel identity into $\operatorname{Range}(W_T)=\mathcal R_T$. [/proofplan]

[step:Compute the Gramian quadratic form as an integral of squares]Fix $T>0$. For $q \in \mathbb{R}^n$, define \begin{align*} g_q : [0,T] \to \mathbb{R}^m, \qquad g_q(s) := B^\top e^{A^\top s}q. \end{align*} The map $g_q$ is continuous because the matrix exponential $s \mapsto e^{A^\top s}$ is continuous. By the definition of $W_T$ and associativity of matrix multiplication, \begin{align*} q^\top W_T q = \int_0^{\!T}q^\top e^{As} B B^\top e^{A^\top s}q\, d\mathcal{L}^1(s). \end{align*} For each $s \in [0,T]$, \begin{align*} q^\top e^{As} B B^\top e^{A^\top s}q = |B^\top e^{A^\top s}q|^2 = |g_q(s)|^2. \end{align*} Therefore \begin{align*} q^\top W_T q = \int_0^{\!T}|g_q(s)|^2\, d\mathcal{L}^1(s). \end{align*}[/step]

[guided]Fix $T>0$ and take an arbitrary vector $q \in \mathbb{R}^n$. We introduce the continuous map \begin{align*} g_q : [0,T] \to \mathbb{R}^m, \qquad g_q(s) := B^\top e^{A^\top s}q. \end{align*} This is the right quantity to isolate because the Gramian has the form $e^{As}BB^\top e^{A^\top s}$, so pairing it with $q$ on both sides should produce the squared Euclidean norm of $B^\top e^{A^\top s}q$. Using the definition of $W_T$, we compute \begin{align*} q^\top W_T q = q^\top \left(\int_0^{\!T}e^{As} B B^\top e^{A^\top s}\, d\mathcal{L}^1(s)\right) q. \end{align*} Since the integrand is a continuous matrix-valued function on the compact interval $[0,T]$, the integral is entrywise well-defined, and multiplication by the fixed vectors $q^\top$ and $q$ may be brought inside the finite-dimensional integral. Hence \begin{align*} q^\top W_T q = \int_0^{\!T}q^\top e^{As} B B^\top e^{A^\top s}q\, d\mathcal{L}^1(s). \end{align*} Now use $(e^{As})^\top=e^{A^\top s}$, so $q^\top e^{As}B=(B^\top e^{A^\top s}q)^\top$. Thus, for each $s \in [0,T]$, \begin{align*} q^\top e^{As} B B^\top e^{A^\top s}q = (B^\top e^{A^\top s}q)^\top(B^\top e^{A^\top s}q)=|B^\top e^{A^\top s}q|^2. \end{align*} Therefore the quadratic form of the Gramian is exactly \begin{align*} q^\top W_T q = \int_0^{\!T}|B^\top e^{A^\top s}q|^2\, d\mathcal{L}^1(s). \end{align*} This identity is the bridge between algebraic information about $W_T$ and dynamical information about which directions the input matrix $B$ can reach.[/guided]

[step:Characterize the kernel of $W_T$ by pointwise annihilation] For each $s \in [0,T]$, the matrix $e^{As}BB^\top e^{A^\top s}$ is symmetric because \begin{align*} \left(e^{As}BB^\top e^{A^\top s}\right)^\top=e^{As}BB^\top e^{A^\top s}. \end{align*} Therefore the entrywise integral $W_T$ is symmetric. If $q \in \ker W_T$, then $q^\top W_Tq=0$, and the quadratic-form identity gives \begin{align*} \int_0^{\!T}|g_q(s)|^2\, d\mathcal{L}^1(s)=0. \end{align*} The function $s \mapsto |g_q(s)|^2$ is continuous and nonnegative. If it were positive at some $s_0 \in [0,T]$, continuity would give an interval of positive $\mathcal{L}^1$-measure on which it is positive, contradicting the vanishing integral. Thus $g_q(s)=0$ for every $s \in [0,T]$. Conversely, if $g_q(s)=0$ for every $s \in [0,T]$, then \begin{align*} W_Tq=\int_0^{\!T}e^{As}B\,g_q(s)\, d\mathcal{L}^1(s)=0. \end{align*} Therefore \begin{align*} q \in \ker W_T \iff B^\top e^{A^\top s}q=0 \text{ for every } s \in [0,T]. \end{align*} [/step]

[step:Identify vectors orthogonal to every reachable terminal state] Let $q \in \mathbb{R}^n$. For each control \begin{align*} u \in L^2((0,T), \mathcal{B}((0,T)), \mathcal{L}^1; \mathbb{R}^m), \end{align*} define the reachable terminal state \begin{align*} x_u := \int_0^{\!T}e^{A(T-s)}Bu(s)\, d\mathcal{L}^1(s) \in \mathbb{R}^n. \end{align*} Then \begin{align*} q^\top x_u = \int_0^{\!T}q^\top e^{A(T-s)}Bu(s)\, d\mathcal{L}^1(s). \end{align*} For each $s \in [0,T]$, \begin{align*} q^\top e^{A(T-s)}Bu(s) = \left(B^\top e^{A^\top(T-s)}q\right)^\top u(s). \end{align*} Thus \begin{align*} q^\top x_u = \int_0^{\!T}\left(B^\top e^{A^\top(T-s)}q\right)^\top u(s)\, d\mathcal{L}^1(s). \end{align*} If $q \in \ker W_T$, the previous step gives $B^\top e^{A^\top r}q=0$ for every $r \in [0,T]$. Taking $r=T-s$ gives $q^\top x_u=0$ for every admissible $u$, so $q \in \mathcal R_T^\perp$. Conversely, suppose $q \in \mathcal R_T^\perp$. Define \begin{align*} u_q : (0,T) \to \mathbb{R}^m, \qquad u_q(s) := B^\top e^{A^\top(T-s)}q. \end{align*} The map $u_q$ is continuous on $(0,T)$ and bounded because it extends continuously to $[0,T]$, hence $u_q \in L^2((0,T), \mathcal{B}((0,T)), \mathcal{L}^1; \mathbb{R}^m)$. Since $q$ is orthogonal to every reachable terminal state, applying this to $x_{u_q}$ gives \begin{align*} 0=q^\top x_{u_q}=\int_0^{\!T}|B^\top e^{A^\top(T-s)}q|^2\, d\mathcal{L}^1(s). \end{align*} The substitution $r=T-s$ preserves one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on the interval and sends $s \in [0,T]$ to $r \in [0,T]$, so \begin{align*} \int_0^{\!T}|B^\top e^{A^\top(T-s)}q|^2\, d\mathcal{L}^1(s)=\int_0^{\!T}|B^\top e^{A^\top r}q|^2\, d\mathcal{L}^1(r). \end{align*} Therefore $q \in \ker W_T$. We have proved \begin{align*} \ker W_T=\mathcal R_T^\perp. \end{align*} [/step]

[step:Pass from the kernel identity to the range identity] Because $W_T$ is symmetric, its range is the orthogonal complement of its kernel. Indeed, if $x \in \mathbb{R}^n$, $y=W_Tx$, and $q \in \ker W_T$, then \begin{align*} q^\top y=q^\top W_Tx=(W_Tq)^\top x=0, \end{align*} so $\operatorname{Range}(W_T)\subseteq(\ker W_T)^\perp$. Both subspaces have dimension equal to $\operatorname{rank}(W_T)$ by rank-nullity, hence equality holds: \begin{align*} \operatorname{Range}(W_T)=(\ker W_T)^\perp. \end{align*} The reachable set $\mathcal R_T$ is a subspace of $\mathbb{R}^n$ because it is the image of the [linear map](/page/Linear%20Map) sending an admissible control to its terminal state. Hence, in finite-dimensional Euclidean space, $(\mathcal R_T^\perp)^\perp=\mathcal R_T$. Using $\ker W_T=\mathcal R_T^\perp$, we obtain \begin{align*} \operatorname{Range}(W_T)=\mathcal R_T. \end{align*} Finally, $W_T$ is nonsingular if and only if $\operatorname{Range}(W_T)=\mathbb{R}^n$. By the range identity, this is equivalent to $\mathcal R_T=\mathbb{R}^n$. This proves both assertions. [/step]