A couple of questions from the Wikipedia "matrix exponential" article:

  1. In the part of the article I linked to, they mention that to conclude that every matrix in $GL(n)$ has a logarithm (though not unique), it is necessary to pass to the complex numbers. Based on this comment, I tried to choose a real matrix (e.g. $-I$) and show that it did not have a real logarithm. I found it difficult, though. Any idea on this proof?
  2. They mention the inequality $\|e^{X+Y}-e^X\| \leq \|Y\| e^{\|X\|} e^{\|Y\|}$. Does anyone have a reference for this proof, or can someone indicate the idea of the proof?
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    $\begingroup$ Hello. Please always include some English words in the title, so that right-clicks are not overridden by MathJax's context menu. $\endgroup$ – user1551 Nov 12 '13 at 18:49
  1. The negative identity matrix does have real logarithms when it is even-dimensional. For instance, $\exp\pmatrix{0&-(2k+1)\pi\\ (2k+1)\pi&0}=-I$ for every integer $k$. It has no real logarithm when it is odd-dimensional, because $e^\lambda=-1$ has not any real solution and nonreal eigenvalues of a real matrix must occur in conjugate pairs.
  2. Proofs of this statement can be found in several books. To begin, note that $\| (X+Y)^n - X^n \|\le(\|X\|+\|Y\|)^n - \|X\|^n$ for each $n$.
  • $\begingroup$ Thank you. Any particular references? I have the books by Strang, Lax, Friedberg, Roman $\endgroup$ – user88203 Nov 14 '13 at 15:03
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    $\begingroup$ @user88203 The inequality is mentioned in corollary 6.2.32 on p.430 of Horn and Johnson's Topics in Matrix Analysis. Since this is a direct consequence of theorem 6.1.10 in the book, the authors have not provided an independent proof of the inequality. Anyway, once you have shown that $\| (X+Y)^n - X^n \|\le(\|X\|+\|Y\|)^n - \|X\|^n$, it follows that $\|e^{X+Y}-e^X\|\le e^{\|X\|+\|Y\|}-e^{\|X\|}$ and the rest is single-variable calculus. $\endgroup$ – user1551 Nov 14 '13 at 16:09

For all commuting matrix $X$, $Y$,

$\|e^{X+Y}-e^X\| = \|e^{X}(e^{Y}-I_{n})\|$

By sub-multiplicativity of the norm $\|.\|$,

$\|e^{X}(e^{Y}-I_{n})\| \leq \|e^{X}\|.\|e^{Y}-I_{n}\|$

Then by definition, $\|e^{Y}-I_{n}\|=\displaystyle\|\sum_{n=0}^\infty\frac{Y^n}{n!}-I_{n}\|=\|\displaystyle\sum_{n=1}^\infty\frac{Y^{n}}{n!}\|=\|Y\displaystyle\sum_{n=1}^\infty\frac{Y^{n-1}}{n!}\|$

Again, by sub-multiplicativity of the norm and triangular inequality,

$\|Y\displaystyle\sum_{n=1}^\infty\frac{Y^{n-1}}{n!}\| \leq\|Y\|.\|\displaystyle\sum_{n=1}^\infty\frac{Y^{n-1}}{n!}\|\leq\|Y\|.\displaystyle\sum_{n=1}^\infty\frac{\|Y\|^{n-1}}{n!}\leq\|Y\|.\displaystyle\sum_{n=1}^\infty\frac{\|Y\|^{n-1}}{(n-1)!}= \|Y\| e^{\|Y\|}$

Eventually, $\|e^{X+Y}-e^X\| \leq \|Y\| e^{\|X\|} e^{\|Y\|}$


Hint: You can consider $e^A=\displaystyle\sum_{n=0}^\infty\frac{A^n}{n!}$ and $\|A^n\|\leq \|A\|^n$ to show that $\|e^{A}\|\leq e^{\|A\|}$.


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