Prove that if $\|A\|<1$, then $\|(I-A)^{-1}\|\geq {1\over1+\|A\|}$.
I'm not sure how to prove this result. I see feel like a geometric series is involved though. Any solutions or hints are appreciated.
2026-03-28 09:33:33.1774690413
Prove that if $\|A\|<1$, then $\|(I-A)^{-1}\|\geq {1\over1+\|A\|}$.
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We know that for $\|A\| < 1$, $(I-A)^{-1}$ is well-defined (prove this yourself if you have not done so yet) so we can talk about the inverse. Thus:
$$ I = (I-A)(I-A)^{-1}.$$
Here is a hint:
$$ 1 = \|I\| = \|(I-A)(I-A)^{-1}\|.$$
Try doing some basic norm manipulations to this. You need to increase the norm, not decrease it since you want a lower bound. (Luiz' comment might be useful here...)
The explicit formula for $(I-A)^{-1}$ is not super useful here (in my opinion) since that will give you an upper bound on the norm, not a lower bound.