Weak Existence of Matrix Logarithm

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Theorem

Let $T$ be a square matrix of order $n$.

Let $\norm {T - I} < 1$ in the norm on bounded linear operators, where $I$ the identity matrix.


Then there is a square matrix $S$ such that:

$e^S = T$

where $e^S$ is the matrix exponential.


Proof

Define:

$\ds S = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n$

$S$ converges since $\norm {T - I} < 1$.

We have that $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \norm {T - I}^n$ is the Newton-Mercator Series.

This converges since $\norm {T - I} < 1$.

Hence the series for $S$ converges absolutely, and so $S$ is well defined.

Using the series definition for the matrix exponential:

\(\ds e^S\) \(=\) \(\ds I + S + \frac 1 {2!} S^2 + \frac 1 {3!} S^3 + \cdots\)
\(\ds \) \(=\) \(\ds I + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n + \frac 1 {2!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n}^2 + \frac 1 {3!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n}^3 + \cdots\)
\(\ds \leadsto \ \ \) \(\ds e^S\) \(=\) \(\ds I + \paren {T - I} + c_2 \paren {T - I}^2 + c_3 \paren {T - I}^3 + \cdots\) grouping terms by powers of $T - I$
\(\ds \) \(=\) \(\ds T + c_2 \paren {T - I}^2 + c_3 \paren {T - I}^3 + \cdots\)

If $c_i = 0$ for $i \ge 2$, then $e^S = T$, and the result is shown.


The Newton-Mercator Series is a Taylor expansion for $\map \ln {1 + x}$.

When combined with the Power Series Expansion for Exponential Function, it gives:

\(\ds e^{\map \ln {1 + x} }\) \(=\) \(\ds 1 + \map \ln {1 + x} + \frac 1 {2!} \paren {\map \ln {1 + x} }^2 + \frac 1 {3!} \paren {\map \ln {1 + x} }^3 + \cdots\)
\(\ds \) \(=\) \(\ds 1 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n + \frac 1 {2!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n}^2 + \frac 1 {3!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n}^3 + \cdots\)
\(\ds \) \(=\) \(\ds 1 + x + c_2 x^2 + c_3 x^3 + \cdots\) grouping terms by powers of $x$

But $e^{\map \ln {1 + x} } = 1 + x$.

Thus:

$1 + x = 1 + x + c_2 x^2 + c_3 x^3 + \cdots \implies c_i = 0$

for $i \ge 2$.

$\blacksquare$


Also see


Sources

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