Finite-dimensional matrix exponential

Let $d\ge 1$. Regard ${\mathbb{C}}^d$ as column vectors with the standard complex inner product $⟨x,y⟩:={\sum }_{j=1}^d{x}_j\overline{y_j},$ linear in the first argument, and induced norm $\mathrm{\parallel }x\mathrm{\parallel }=({\sum }_{j=1}^d\mathrm{\mid }{x}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. A matrix $A=(a_{ij})\in M_d(\mathbb{C})$ acts by $(Ax{)}_i={\sum }_{j=1}^d{a}_{ij}{x}_j$, and its adjoint for this inner product is $A^{\ast }={\bar{A}}^{\mathsf{T}}$.

For $A\in M_d(\mathbb{C})$ and $z\in \mathbb{C}$, define the finite-dimensional matrix exponential by $e^{zA}:={\sum }_{k=0}^{\mathrm{\infty }}\frac{z^k{A}^k}{k!},A^0=I_d\mathrm{.}$ The scalar exponential convention is $\exp w={\sum }_{k\ge 0}{w}^k\mathrm{/}k!$.

Every matrix $B=(b_{ij})\in M_d(\mathbb{C})$ is bounded on ${\mathbb{C}}^d$: since $\mathrm{\mid }{x}_j\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$, one has $\mathrm{\parallel }Bx{\mathrm{\parallel }}^2\le {\sum }_{i=1}^d{\left({\sum }_{j=1}^d\mathrm{\mid }{b}_{ij}\mathrm{\mid }\right)}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{.}$ Thus its operator norm is defined by the unit-ball supremum in the operator-norm definition listed in Definition references. For $B,C\in M_d(\mathbb{C})$ and $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, the operator-norm bound gives $\mathrm{\parallel }BCx\mathrm{\parallel }\le \mathrm{\parallel }B\mathrm{\parallel }\mathrm{\parallel }Cx\mathrm{\parallel }\le \mathrm{\parallel }B\mathrm{\parallel }\mathrm{\parallel }C\mathrm{\parallel }$; taking the supremum proves $\mathrm{\parallel }BC\mathrm{\parallel }\le \mathrm{\parallel }B\mathrm{\parallel }\mathrm{\parallel }C\mathrm{\parallel }$. Since $d\ge 1$, the coordinate vector $e_1$ with first entry $1$ and other entries $0$ is a unit vector, so $\mathrm{\parallel }{I}_d\mathrm{\parallel }=1$. Homogeneity and the triangle inequality follow from the unit-ball supremum, and $\mathrm{\parallel }B\mathrm{\parallel }=0$ forces $B{e}_j=0$ for each coordinate vector $e_j$ (with entry $1$ at $j$ and $0$ elsewhere), hence $B=0$. Therefore $M_d(\mathbb{C})$ with this operator norm is a normed $\mathbb{C}$-vector space; the matrices $E_{ij}$ with entry $1$ at $(i,j)$ and $0$ elsewhere form a basis of size $d^2$.

Completeness of the complex plane and finite-dimensional norm equivalence (the two theorems in Definition references) make $M_d(\mathbb{C})$ complete in this operator norm. By induction and submultiplicativity, $\mathrm{\parallel }{A}^k\mathrm{\parallel }\le \mathrm{\parallel }A{\mathrm{\parallel }}^k$ for every $k\ge 1$, while $\mathrm{\parallel }{A}^0\mathrm{\parallel }=1$. The scalar series ${\sum }_{k\ge 0}(\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }A\mathrm{\parallel }{)}^k\mathrm{/}k!$ converges by the scalar exponential convergence lemma in Definition references. Hence for $m>n$, $\parallel {\sum }_{k=n+1}^m\frac{z^k{A}^k}{k!}\parallel \le {\sum }_{k=n+1}^m\frac{(\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }A\mathrm{\parallel }{)}^k}{k!}⟶0(n\to \mathrm{\infty })\mathrm{.}$ The partial sums are Cauchy, so completeness gives their operator-norm limit; the same estimate proves absolute convergence in operator norm.

The complex exponential by its power series

The operator norm as the least bound and as the unit-sphere or unit-ball supremum

Real and complex inner-product spaces and their induced length

The complex exponential series converges absolutely for every complex argument

The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts

Finite dimensional norm equivalence over a complete valued field