Proof of $(I-C)^{-1}\leq \|I\| + \|C\| + \|C\|^2 + \cdots$ Using Matrix C

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SUMMARY

The discussion focuses on proving the inequality \(\|(I - C)^{-1}\| \leq \|I\| + \|C\| + \|C\|^2 + \cdots\) under the condition that the matrix \(I - C\) is invertible. The participants draw a parallel between this inequality and the geometric series formula \(\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots\). The proof method suggested involves leveraging properties of matrix norms and the series expansion, confirming the validity of the inequality.

PREREQUISITES
  • Understanding of matrix norms and their properties
  • Knowledge of matrix inversion and invertibility conditions
  • Familiarity with geometric series and convergence criteria
  • Basic linear algebra concepts, including matrices and operations
NEXT STEPS
  • Study the properties of matrix norms in detail
  • Learn about the conditions for matrix invertibility, specifically for \(I - C\)
  • Explore the derivation and applications of geometric series in linear algebra
  • Investigate further proofs related to matrix inequalities and their implications
USEFUL FOR

Mathematicians, students of linear algebra, and researchers interested in matrix theory and inequalities will benefit from this discussion.

flop
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Given that [itex]C[/itex] is a matrix and that [itex]I - C[/itex] is invertible, could someone show me the following?

[itex]\|(I - C)^{-1}\| \leq \|I\| + \|C\| + \|C\|^2 + \cdots[/itex]
 
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That formula looks awfully similar to:

[tex] \frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots[/tex]

Maybe the method of proof is similar?
 
got it! thank you hurky
 

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