Proof of Existence and Uniqueness of the State-Transition Matrix of a Linear System

By Basic Facts about Intervals of the Real Line (v), $[a,b]$ is an interval containing the two distinct points $a$ and $b$; it is the interval $I$ of Rules for Derivatives of Vector- and Matrix-Valued Maps and The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps throughout.

Step 1: two auxiliary solutions. Let $O:[a,b]\to\mathcal{M}_{n}(\mathbb{R})$ be the constant map with value $0_{n}$ and $N:[a,b]\to\mathcal{M}_{n}(\mathbb{R})$ the map $t\mapsto(-I_{n})A(t)=-A(t)$ (by Products of Real Matrices with Zero Matrices, Negatives and Differences (ii) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv)). Both are continuous on $[a,b]$ by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v), with $(\mathbb{R},d_{\mathbb{R}})$ for its $(X,d)$ and $[a,b]$ for its $S$: $O$ is constant, and $N$ is the product of a constant map and $A$. By Linear Systems with Continuous Coefficients: Continuity, Global Lipschitz Bound and Well-Posedness of the Initial Value Problem (i) and Linear Systems with Continuous Coefficients: Continuity, Global Lipschitz Bound and Well-Posedness of the Initial Value Problem (v), with $n$ for $q$ and $A,O,O$ for $A,B,G$, the right-hand side is the map $(t,Y)\mapsto A(t)Y+Y0_{n}+0_{n}=A(t)Y$, by Products of Real Matrices with Zero Matrices, Negatives and Differences (i) (and Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (i)); let $U$ be the solution with initial value $I_{n}$. With $O,N,O$ for $A,B,G$, the right-hand side is the map $(t,Y)\mapsto0_{n}Y+Y(-A(t))+0_{n}=-\bigl(YA(t)\bigr)$, by Products of Real Matrices with Zero Matrices, Negatives and Differences (i) and Products of Real Matrices with Zero Matrices, Negatives and Differences (ii) (and Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (i)); let $V$ be the solution with initial value $I_{n}$. By Solution of an Initial Value Problem on $[a,c]$ (iii) and Solution of an Initial Value Problem on $[a,c]$ (ii), with $b$ for $c$, $U,V:[a,b]\to\mathcal{M}_{n}(\mathbb{R})$ are differentiable and, for every $t\in[a,b]$,

$$U'(t)=A(t)U(t),\qquad V'(t)=-\bigl(V(t)A(t)\bigr),\qquad U(a)=V(a)=I_{n}.$$

Step 2: $V(t)U(t)=I_{n}=U(t)V(t)$ for every $t\in[a,b]$. Let $P:[a,b]\to\mathcal{M}_{n}(\mathbb{R})$, $t\mapsto V(t)U(t)$, be the pointwise product. By Rules for Derivatives of Vector- and Matrix-Valued Maps (v), $P$ is differentiable at every $t\in[a,b]$, and, by Products of Real Matrices with Zero Matrices, Negatives and Differences (ii) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i) (and Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (i)),

$$P'(t)=V'(t)U(t)+V(t)U'(t)=-\bigl(V(t)A(t)\bigr)U(t)+V(t)\bigl(A(t)U(t)\bigr)=0_{n}.$$

So $P$ is differentiable (Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (iii)), and by The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (iv) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv), $P(t)=P(a)=I_{n}I_{n}=I_{n}$ for every $t\in[a,b]$. Let $t\in[a,b]$. As $V(t)U(t)=I_{n}$, $U(t)$ is invertible by Criteria for Invertibility of a Square Real Matrix (v) and Criteria for Invertibility of a Square Real Matrix (vi). By Inverse Matrix and Invertible Real Square Matrix (i), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv),

$$V(t)=V(t)\bigl(U(t)U(t)^{-1}\bigr)=\bigl(V(t)U(t)\bigr)U(t)^{-1}=U(t)^{-1},$$

so $U(t)V(t)=U(t)U(t)^{-1}=I_{n}$.

Step 3: proof of (i). Let $s$, $q$ and $X$ be as there, and let $Q:[a,b]\to\mathcal{M}_{n\times q}(\mathbb{R})$, $t\mapsto V(t)X(t)$. By Rules for Derivatives of Vector- and Matrix-Valued Maps (v), Products of Real Matrices with Zero Matrices, Negatives and Differences (ii) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i) (and Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (i)), $Q$ is differentiable at every $t\in[a,b]$ with

$$Q'(t)=V'(t)X(t)+V(t)X'(t)=-\bigl(V(t)A(t)\bigr)X(t)+V(t)\bigl(A(t)X(t)\bigr)=0_{n\times q}.$$

By The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (iv) and Products of Real Matrices with Zero Matrices, Negatives and Differences (i), $Q(t)=Q(s)=V(s)0_{n\times q}=0_{n\times q}$ for every $t\in[a,b]$. Hence, by Step 2, Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i) and Products of Real Matrices with Zero Matrices, Negatives and Differences (i),

$$X(t)=\bigl(U(t)V(t)\bigr)X(t)=U(t)Q(t)=U(t)0_{n\times q}=0_{n\times q}\qquad\bigl(t\in[a,b]\bigr).$$

For the vector case, let $x$ be as there and $w:[a,b]\to\mathbb{R}^{n}$, $t\mapsto V(t)x(t)$. By Rules for Derivatives of Vector- and Matrix-Valued Maps (vi), Products of Real Matrices with Zero Matrices, Negatives and Differences (ii) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v) (and Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (i)), $w$ is differentiable at every $t\in[a,b]$ with

$$w'(t)=V'(t)x(t)+V(t)x'(t)=-\bigl(V(t)A(t)\bigr)x(t)+V(t)\bigl(A(t)x(t)\bigr)=0 .$$

By The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (iv) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v), $w(t)=w(s)=V(s)0=0$, and so $x(t)=I_{n}x(t)=\bigl(U(t)V(t)\bigr)x(t)=U(t)w(t)=U(t)0=0$ for every $t\in[a,b]$. This proves (i).

Step 4: proof of (ii). Existence. Let $\Phi:[a,b]\times[a,b]\to\mathcal{M}_{n}(\mathbb{R})$ be the map $(t,s)\mapsto U(t)V(s)$ (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii), with the matrix product for its $F$). Let $s\in[a,b]$. The map $\Phi(\cdot,s)$ is $t\mapsto\bigl(I_{n}U(t)\bigr)V(s)$, by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv). By Rules for Derivatives of Vector- and Matrix-Valued Maps (x), with $I_{n}$ for $M$ and $V(s)$ for $N$, it is differentiable at every $t\in[a,b]$, hence differentiable, with derivative

$$\bigl(I_{n}U'(t)\bigr)V(s)=\bigl(A(t)U(t)\bigr)V(s)=A(t)\Phi(t,s),$$

by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i); and $\Phi(s,s)=U(s)V(s)=I_{n}$ by Step 2.

Uniqueness. Let $\tilde{\Phi}:[a,b]\times[a,b]\to\mathcal{M}_{n}(\mathbb{R})$ have the same properties, and let $s\in[a,b]$. The map $D:[a,b]\to\mathcal{M}_{n}(\mathbb{R})$, $t\mapsto\Phi(t,s)-\tilde{\Phi}(t,s)$, is differentiable by Rules for Derivatives of Vector- and Matrix-Valued Maps (iv), with

$$D'(t)=A(t)\Phi(t,s)-A(t)\tilde{\Phi}(t,s)=A(t)D(t)\qquad\bigl(t\in[a,b]\bigr)$$

by Products of Real Matrices with Zero Matrices, Negatives and Differences (iii), and $D(s)=I_{n}-I_{n}=0_{n}$ (Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (ii)). By (i), with $n$ for $q$ and $D$ for $X$, $D(t)=0_{n}$, that is, $\Phi(t,s)=\tilde{\Phi}(t,s)$, for every $t\in[a,b]$. As $s$ was arbitrary, $\Phi$ and $\tilde{\Phi}$ have the same values, so $\Phi=\tilde{\Phi}$. This proves (ii).