Weak Existence of Matrix Logarithm

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

Let $\left\Vert{T - I}\right\Vert < 1$ in the operator norm, 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:


 * $\displaystyle S = \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n - 1} } n \left({T - I}\right)^n$

$S$ converges since $\left\Vert{T - I}\right\Vert < 1$. We have that $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n - 1} } n \left\Vert{T - I}\right\Vert^n$ is the Newton-Mercator Series and converges since $\left\Vert{T - I}\right\Vert < 1$. Hence the series for $S$ converges absolutely, and so $S$ is well defined.

Using the series definition for the matrix exponential:

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 $\ln \left({1 + x}\right)$.

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

But $e^{\ln \left({1 + x}\right)} = 1 + x$.

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

for $i \ge 2$.

Also see

 * Existence of Matrix Logarithm