Proof of Properties of the Matrix Exponential

Computations with real numbers use The Real Numbers as a Type. $(\mathcal{M}_{n}(\mathbb{R}),d_{n})$ is a metric space by The Real Matrices with the Operator-Norm Distance Form a Complete Metric Space with Entrywise Convergence (i), with $d_{n}(X,Y)=\lVert X-Y\rVert$ (Euclidean Space and Real Matrices: Standing Notation (vi)). For $\gamma\in\mathbb{R}$ with $\gamma>0$ and $X\in\mathcal{M}_{n}(\mathbb{R})$, let $C_{\gamma,X}:[-\gamma,\gamma]\to\mathcal{M}_{n}(\mathbb{R})$ be the constant map with value $X$. Then $-\gamma<\gamma$, and $C_{\gamma,X}$ is continuous on $[-\gamma,\gamma]$ 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). By The State-Transition Matrix of a Constant Coefficient Matrix Is a Matrix Exponential (i), with $-\gamma$, $\gamma$ and $X$ in place of $a$, $b$ and $M$,

$$\Phi_{C_{\gamma,X}}(r,q)=e^{(r-q)X}\qquad(q,r\in[-\gamma,\gamma]).\tag{$\ast$}$$

Proof of (i). By $(\ast)$ with $\gamma=1$, $X=M$ and $q=r=0$, and by The State-Transition Matrix of a Linear System (iii), $e^{0M}=\Phi_{C_{1,M}}(0,0)=I_{n}$. As $0M=0_{n}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv), $e^{0_{n}}=I_{n}$.

Proof of (ii). Let $i,j\in[n]$. For $k\in\mathbb{N}$, $(M^{k-1})_{ji}=\bigl((M^{k-1})^{\top}\bigr)_{ij}=\bigl((M^{\top})^{k-1}\bigr)_{ij}$ by Transpose of a Real Matrix (i) and Powers of a Square Matrix (v); so the sequences $k\mapsto(M^{k-1})_{ji}/(k-1)!$ and $k\mapsto((M^{\top})^{k-1})_{ij}/(k-1)!$ have the same terms and are equal, and so are the sums of their series. By Transpose of a Real Matrix (i) and The Exponential of a Real Square Matrix (i), applied to $M$ and to $M^{\top}$,

$$\bigl((e^{M})^{\top}\bigr)_{ij}=(e^{M})_{ji}=\sum_{k=1}^{\infty}\frac{(M^{k-1})_{ji}}{(k-1)!}=\sum_{k=1}^{\infty}\frac{((M^{\top})^{k-1})_{ij}}{(k-1)!}=(e^{M^{\top}})_{ij}.$$

As $i,j$ were arbitrary, $(e^{M})^{\top}=e^{M^{\top}}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iii).

Proof of (iii). Let $B\in\mathcal{M}_{n}(\mathbb{R})$ with $BM=MB$, and for $N\in\mathbb{N}$ let $S_{N}=S_{N}(M)$ be the matrix of The Exponential Series of a Real Square Matrix: Convergence, a Norm Bound and the Exponential Map; $(S_{N})_{N\in\mathbb{N}}$ is a sequence in $\mathcal{M}_{n}(\mathbb{R})$ by The Exponential Series of a Real Square Matrix: Convergence, a Norm Bound and the Exponential Map (i), and so are $(BS_{N})_{N\in\mathbb{N}}$ and $(S_{N}B)_{N\in\mathbb{N}}$, the maps $N\mapsto BS_{N}$ and $N\mapsto S_{N}B$ formed from it and the constant sequence with value $B$ with the matrix product (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii)). Here $S_{1}=I_{n}$ and $S_{N+1}=S_{N}+\frac{1}{N!}M^{N}$ by The Exponential Series of a Real Square Matrix: Convergence, a Norm Bound and the Exponential Map (iii). We show $BS_{N}=S_{N}B$ for every $N\in\mathbb{N}$ by induction on $N$ (Arithmetic and Order of the Natural Numbers (xiii)). For $N=1$, $S_{1}=I_{n}$ and $BI_{n}=B=I_{n}B$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv). If $BS_{N}=S_{N}B$, then by The Exponential Series of a Real Square Matrix: Convergence, a Norm Bound and the Exponential Map (iii), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (ii), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iii) and Powers of a Square Matrix (iii),

$$BS_{N+1}=BS_{N}+\frac{1}{N!}(BM^{N})=S_{N}B+\frac{1}{N!}(M^{N}B)=\Bigl(S_{N}+\frac{1}{N!}M^{N}\Bigr)B=S_{N+1}B .$$

The constant sequence with value $B$ converges to $B$ in $(\mathcal{M}_{n}(\mathbb{R}),d_{n})$ by Convergent Sequence in a Metric Space (i), as $d_{n}(B,B)=\lVert0_{n}\rVert=0$ by Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences (ii) and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (iv); and $(S_{N})$ converges to $e^{M}$ by The Exponential of a Real Square Matrix (ii). By Limits of Matrix Operations, and the Closed Complete Sets of Symmetric and of Positive Semidefinite Matrices (iv), with $n$ in place of $m$ and $p$, the sequence $(BS_{N})$ converges to $Be^{M}$ and the sequence $(S_{N}B)$ converges to $e^{M}B$. These sequences are equal, so $Be^{M}=e^{M}B$ by Uniqueness of Limits in a Metric Space (i).

Proof of (iv). Let $s,t\in\mathbb{R}$ and $\gamma=|s|+|t|+|s+t|+1$, so $\gamma>0$ and $0$, $t$ and $s+t$ lie in $[-\gamma,\gamma]$. By $(\ast)$ with $X=M$, and by Basic Properties of the State-Transition Matrix (ii) for $C_{\gamma,M}$,

$$e^{(s+t)M}=\Phi_{C_{\gamma,M}}(s+t,0)=\Phi_{C_{\gamma,M}}(s+t,t)\,\Phi_{C_{\gamma,M}}(t,0)=e^{sM}e^{tM},$$

as $(s+t)-0=s+t$, $(s+t)-t=s$ and $t-0=t$.

Proof of (v). By (iv) with $1$ and $-1$ in place of $s$ and $t$, and with $-1$ and $1$ in place of $s$ and $t$, and by (i), using $0M=0_{n}$, $1M=M$ and $(-1)M=-M$ (Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv)),

$$e^{M}e^{-M}=e^{0M}=I_{n},\qquad e^{-M}e^{M}=e^{0M}=I_{n}.$$

So $e^{-M}$ is an inverse of $e^{M}$ (Inverse Matrix and Invertible Real Square Matrix (i)), $e^{M}$ is invertible, and $(e^{M})^{-1}=e^{-M}$ by Inverse Matrix and Invertible Real Square Matrix (ii).

Proof of (vi). Let $a<b$. By Basic Facts about Intervals of the Real Line (v), $[a,b]$ is an interval containing two distinct points, and derivatives of maps on $[a,b]$ are taken relative to $[a,b]$. Let $H:[a,b]\to\mathcal{M}_{n}(\mathbb{R})$ be the constant map with value $M$, 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). By The State-Transition Matrix of a Constant Coefficient Matrix Is a Matrix Exponential (i), $\Phi_{H}(u,a)=e^{(u-a)M}$ for $u\in[a,b]$, and by The State-Transition Matrix of a Linear System (ii) the map $u\mapsto\Phi_{H}(u,a)$ is differentiable with derivative $H(u)\Phi_{H}(u,a)=M\Phi_{H}(u,a)$ at $u\in[a,b]$. By (iv) with $a$ and $u-a$ in place of $s$ and $t$,

$$e^{uM}=e^{aM}e^{(u-a)M}=\bigl(e^{aM}\Phi_{H}(u,a)\bigr)I_{n}\qquad(u\in[a,b]),$$

using Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv). For every $w\in\mathbb{R}$, $M(wM)=w(MM)=(wM)M$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iii), so $Me^{wM}=e^{wM}M$ by (iii) with $wM$ and $M$ in place of $M$ and $B$; we use this for $w=a$ and for $w=u\in[a,b]$. By Rules for Derivatives of Vector- and Matrix-Valued Maps (x), with $[a,b]$ in place of $I$ and $e^{aM}$ and $I_{n}$ in place of its $M$ and $N$, the map $u\mapsto e^{uM}$ is differentiable at every $u\in[a,b]$, hence on $[a,b]$ (Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (iii)), with derivative $\bigl(e^{aM}M\Phi_{H}(u,a)\bigr)I_{n}=e^{aM}M\,e^{(u-a)M}=Me^{aM}e^{(u-a)M}=Me^{uM}=e^{uM}M$, by associativity (Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i)) and (iv).

Proof of (vii). Let $A,B\in\mathcal{M}_{n}(\mathbb{R})$ with $AB=BA$. By Basic Facts about Intervals of the Real Line (v), $[0,1]$ is an interval containing two distinct points. Let $G:[0,1]\to\mathcal{M}_{n}(\mathbb{R})$ be the constant map with value $A+B$, continuous on $[0,1]$ 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), and let $X,Y:[0,1]\to\mathcal{M}_{n}(\mathbb{R})$ be the maps $X(u)=e^{uA}e^{uB}$ and $Y(u)=e^{u(A+B)}$.

By (vi) with $0$ and $1$ in place of $a$ and $b$, applied to $A$, to $B$ and to $A+B$ in place of $M$, the maps $u\mapsto e^{uA}$, $u\mapsto e^{uB}$ and $Y$ are differentiable, with derivatives $Ae^{uA}$, $Be^{uB}$ and $(A+B)Y(u)$ at $u\in[0,1]$. For $u\in[0,1]$, $B(uA)=u(BA)=u(AB)=(uA)B$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iii), so $e^{uA}B=Be^{uA}$ by (iii) with $uA$ in place of $M$. By Rules for Derivatives of Vector- and Matrix-Valued Maps (v), with $[0,1]$ in place of $I$, $X$ is differentiable, and by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (ii)

$$X'(u)=(Ae^{uA})e^{uB}+e^{uA}(Be^{uB})=AX(u)+(Be^{uA})e^{uB}=AX(u)+BX(u)=(A+B)X(u)\qquad(u\in[0,1]).$$

By (i), as $0A=0B=0(A+B)=0_{n}$, $X(0)=I_{n}I_{n}=I_{n}$ and $Y(0)=I_{n}$. So $X$ and $Y$ are differentiable maps $[0,1]\to\mathcal{M}_{n}(\mathbb{R})$ with $X'(u)=G(u)X(u)$, $Y'(u)=G(u)Y(u)$ for every $u\in[0,1]$ and $X(0)=Y(0)=I_{n}$. By Basic Properties of the State-Transition Matrix (i), with $0$ and $1$ in place of $a$ and $b$, $G$ in place of $A$, $0$ in place of $s$, $n$ in place of $q$ and $I_{n}$ in place of $\Xi$, each of $X$ and $Y$ is the map $u\mapsto\Phi_{G}(u,0)I_{n}$; so $X=Y$, and in particular $X(1)=Y(1)$. As $1A=A$, $1B=B$ and $1(A+B)=A+B$, this is $e^{A}e^{B}=e^{A+B}$.