1

I need to prove the following inequality

$\log(I-A)<={\|A\|}/({1-\|A\|})$

expansion is same as reals. i tried to apply triangular ineqality but i have a confusion if i can apply it to infinite series or not if yes then its easy .then we can apply ineqality and finally come at a infinite geometric series and that will sum up to write term.

But i am stuck at it ,can somebody help?

biswpo
  • 235
  • How do you define $\log(I-A)$? – copper.hat Mar 25 '14 at 16:14
  • 1
    It seems you'll need to assume $||A||<1$, right? The left hand side is an operator, the rhs is a real number. Do you want $||\log(I-A)||$ on the lhs? – Will Nelson Mar 25 '14 at 18:50
  • Just use the expansion $-\log(I-A) = A + \frac12 A^2 + \frac13 A^3 \dots$, and apply the triangle inequality and the submultiplicity of the norm. Is that what you meant by your proposed approach? Also @WillNelson is correct. – Stephen Montgomery-Smith Mar 26 '14 at 02:41
  • yes $|A|\leq 1$ and it is $|log(I-A)|$ i am just confused is it ok to apply triangle inequality to an infinite series??do i need to assume something about convergence <i am not clear about it – biswpo Mar 26 '14 at 03:05

0 Answers0