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Problem of the Week #209 - May 31, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Define the logarithm of an $n\times n$ matrix $A$ by the power series

$$\sum_{k = 1}^\infty \frac{(-1)^{k-1}(A - I)^k}{k}$$

which converges for $\|A - I\| < 1$ (the standard matrix norm is being used here). Prove that for all $n\times n$ matrices $A$ and $B$ with $\|A - I\| < 1$, $\|B - I\| < 1$, $\|AB - I\| < 1$, and $AB = BA$,

$$\log(AB) = \log A + \log B$$

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
This week's problem was solved correctly by Opalg . You can read his solution below.



Let $D = \{z\in\mathbb{C}: |z-1|<1\}$ and $\exp D = \{e^z:z\in D\}.$ Neither of these sets intersect the negative real axis, so the logarithm is defined as a holomorphic function on both sets, and satisfies the condition \(\displaystyle \log z = \sum_{k=1}^\infty \frac{(-1)^{k-1}(z-1)^k}k\) for $z\in D.$


The holomorphic functional calculus says that given an $n\times n$ matrix $T$ whose spectrum [set of eigenvalues] lies in some domain $U$, there is a continuous unital homomorphism $\phi_T$ from the algebra of holomorphic functions on $U$ (with the topology of uniform convergence on compact subsets) to the $n\times n$ matrices, with $\phi_T(z) = T.$ In particular, if $U = D$ or $U = \exp D$ then this mapping takes the logarithm function to a matrix $\log T$. The continuity of $\phi_T$ ensures that when $U = D$ this definition of $\log T$ agrees with that in the statement of the problem. Also, the exponential and logarithm functions are inverses of each other between the spaces $D$ and $\exp D$.


Given $A$ and $B$ as in the problem, the conditions $\|A - I\|<1$ and $\|B - I\|<1$ ensure that the spectrum of each matrix lies in $D$. Let $X = \log A$ and $Y = \log B.$ Then $YX = XY$, because polynomials in $A$ and $B$ commute. Thus $$\exp(X+Y) = \sum_{n=0}^\infty \frac{(X+Y)^n}{n!} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{X^kY^{n-k}}{k!(n-k)!} = \sum_{s=0}^\infty \frac{X^s}{s!} \sum_{t=0}^\infty \frac{Y^t}{t!} = \exp X \exp Y,$$ the change in order of summation being justified because of the uniform convergence of the series on compact sets.


Therefore $\exp (\log A + \log B) = \exp(X+Y) = \exp X\exp Y = AB$. The condition $\|AB - I\|<1$ ensures that the spectrum of $AB\;(=\exp (\log A + \log B))$ is in $D$, so we can apply the logarithm function to both sides to conclude that $\log A + \log B = \log(AB)$.
 
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