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The logarithm of a matrix $$ \ln(I+A)=\sum_{k=1}^{\infty}{(-1)^{k+1}\over k}A^k$$

converges when ${\rho}(A)<1$

Suppose $n>{\rho}(A)>1$.

Can one use the following transformation $$I+A=(1-n)I + nI + A = (1-n)I +n(I+{A\over n})$$ and then $$\ln(n(I+{A\over n}))+\ln(I+e^{\ln((1-n)I)-\ln(n(I+{A\over n}))})$$ to obtain $$\ln(nI)+\ln(I+{A\over n}) + \ln(I+e^{\ln((1-n)I)-\ln(n(I+{A\over n}))})$$ as the second term now converges?

How can the third term be evaluated?

giorgi
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1 Answers1

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Wouldn't $n \left( I + \frac{1}{n}A \right)$, for which you need a logarithm in the exponent in the third term, have even bigger eigenvalues, hence defeating the purpose of all this?

Since I'm not that familiar with matrix functions, I'll just direct you to the following paper: Al-Mohy, Higham, "Improved Inverse Scaling and Squaring Algorithms for the Matrix Logarithm". The main idea is to use matrix square root to get the eigenvalues of $A^{1/2^s}$ closer to one. I've seen some talks on the subject and the results seem pretty good.

Vedran Šego
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  • What matters is that the spectral radius of ${A\over n}$ decreases so that we arrive at $I{\ln n}+{\sum_{k=1}^{\infty}{(-1)^k\over {n^k \times k}}A^k}$.

    In this case the convergence radius should become n, not 1, as far as I understand. Thank you for the paper.

     – giorgi Aug 01 '13 at 16:27
  • I understand your idea, but you still have $$\ln \left( n \left( I + \frac{1}{n}A \right) \right).$$ I doubt that Al-Mohy and Higham would resort to square root if a simple division would solve this. – Vedran Šego Aug 01 '13 at 16:29
  • Yes, that the power series of $\ln(I+e^{\ln((1-n)I)-\ln(n(I+{A\over n}))})$ should diverge and $Y=e^{\ln((1-n)I)-\ln(n(I+{A\over n}))})$ should have $\rho(Y)>1$ is clear. If the trace of the third term could be evaluated. The paper you referred to is really helpful. – giorgi Aug 01 '13 at 17:40