Let T be a square matrix and I be the corresponding Identity Matrix.

Show that if $\displaystyle \| T-I \|$ is sufficiently small, there exists a linear Operator S such that $\displaystyle e^S = T$.

Hint: expand log(1+x) in a Taylor series.

Printable View

- Mar 15th 2010, 05:05 PMEinStoneExponential of matrices
Let T be a square matrix and I be the corresponding Identity Matrix.

Show that if $\displaystyle \| T-I \|$ is sufficiently small, there exists a linear Operator S such that $\displaystyle e^S = T$.

Hint: expand log(1+x) in a Taylor series. - Mar 17th 2010, 12:59 PMOpalg
Let $\displaystyle V = T-I$, and suppose that $\displaystyle \|V\|<1$. Then define $\displaystyle S = \log(I+V) = V - \frac{V^2}2 + \frac{V^3}3 -\frac{V^4}4 + \ldots$. The series converges absolutely, because $\displaystyle \sum\frac{\|V^n\|}n \leqslant\sum\frac{\|V\|^n}n$ and that series has radius of convergence 1. Then $\displaystyle e^S = \exp\log(I+V) = I+V = T$.