Property for bounding matrix exponential

Question

Wikipedia states in the exponential map section about the exponential of a matrix that for any matrices $X$, $Y$ it holds that $\|e^{X+Y}-e^{X}\| \leq \|Y\|e^{\|X\|} e^{\|Y\|}$ where $\|\cdot\|$ denotes an arbitrary matrix norm. I am trying to prove it since I cannot find it as a reference somewhere. Using the series I can't say that the result is the desired one.

Answer 1

For any $n\times n$ matrices $X_1$ and $X_2$, \begin{equation} \begin{aligned} e^{X_1+X_2}-e^{X_1}&=\sum_{k=0}^\infty\frac1{k!}\,[(X_1+X_2)^k-X_1^k] \\ &=\sum_{k=0}^\infty\frac1{k!}\; \sum_{(j_1,\dots,j_k)\in\{1,2\}^k\setminus\{(1,\dots,1)\}}\; \prod_{i=1}^k X_{j_i}, \end{aligned} \end{equation} so that (assuming that the matrix norm is an operator norm) \begin{equation} \begin{aligned} \|e^{X_1+X_2}-e^{X_1}\| &\le\sum_{k=0}^\infty\frac1{k!}\; \sum_{(j_1,\dots,j_k)\in\{1,2\}^k\setminus\{(1,\dots,1)\}}\; \prod_{i=1}^k \|X_{j_i}\| \\ &=\sum_{k=0}^\infty\frac1{k!}\; \sum_{m=1}^k \binom km \|X_1\|^m \|X_2\|^{k-m} \\ &=\sum_{k=0}^\infty\frac1{k!}\;[(\|X_1\|+\|X_2\|)^k-\|X_1\|^k] \\ &=e^{\|X_1\|+\|X_2\|}-e^{\|X_1\|} \\ &\le \|X_2\| e^{\|X_1\|+\|X_2\|}.\quad\Box \end{aligned} \end{equation}

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As shown in Christian Remling's answer, the condition that the matrix norm is an operator norm cannot be dropped.

Answer 2

The wikipedia article you refer to is here. This indeed states that $$ \|e^{X+Y}-e^X\| \le \|Y\| e^{\|X\|}e^{\|Y\|} \tag{1}\label{1} $$ for any "matrix norm", and, according to the link that is given, this term just means an arbitrary norm on the vector space of matrices, not necessarily satisfying $\|AB\| \le \| A\|\,\|B\|$.

In this generality, the statement is false. Take $$ X=0, \quad Y=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} . $$ Then $e^Y =\cosh 1 + Y\sinh 1$, so if we define a norm as $$ \left\| \begin{pmatrix} a & b \\ c & d \end{pmatrix} \right\| = |a|+|d|+\epsilon (|b| + |c|) $$ and take $\epsilon>0$ small, then the RHS of \eqref{1} will be small while the LHS is $\ge 2(\cosh 1 -1)$.