Analytic function of a matrix: Difference between revisions

Revision as of 17:17, 2 March 2021

Revision as of 18:54, 13 March 2021

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* Using scalar Taylor expansion for <math>f(a+\eta b)</math> and replacing scalars with matrices at the end :

:: <math>\begin{array}{rclalign}

f(a+\eta b) &=& f(a) + f'(a)\frac{\eta b}{1!} + f''(a)\frac{(\eta b)^{2}}{2!} + f'''(a)\frac{(\eta b)^{3}}{3!} \\[.5em]

&=& a^{3} + 3a^{2}(\eta b) + 3a(\eta b)^{2} + (\eta b)^{3} \\[.5em]

&\to & A^{3} = + 3A^{2}(\eta B) + 3A(\eta B)^{2} + (\eta B)^{3}

\end{arrayalign}</math>

The scalar expression assumes commutativity while the matrix expression does not, and thus they cannot be equated directly unless <math>[A,B]=0</math>. For some ''f''(''x'') this can be dealt with using the same method as scalar Taylor series. For example, <math display="inline">f(x) = \frac{1}{x}</math>. If <math>A^{-1}</math> exists then <math>f(A+\eta B) = f(\mathbb{I} + \eta A^{-1}B)f(A)</math>. The expansion of the first term then follows the power series given above,

:<math>f(\mathbb{I} + \eta A^{-1}B) = \mathbb{I} - \eta A^{-1}B + (-\eta A^{-1}B)^{2} + \cdots = \sum_{n=0}^{\infty} (-\eta A^{-1}B)^{n} </math>

The convergence criteria of the power series then apply, requiring <math>\Vert \eta A^{-1}B \Vert</math> to be sufficiently small under the appropriate matrix norm. For more general problems, which cannot be rewritten in such a way that the two matrices commute, the ordering of matrix products produced by repeated application of the Leibniz rule must be tracked.