5

On some online forum, there is the claim:

Given some square matrix:

$$(I - A)^{-1} = I + A + A^2 + A^3 + \ldots$$

This is true if the right side converges, which is true if and only if all of the eigenvalues of A have absolute value smaller than $1$.

Reference https://www.physicsforums.com/threads/matrix-inverse-equals-power-series.423897/

I really like this result, because it relies on the more intuitive spectral radius rather than matrix norm which is defined as:

\begin{align} \|A\| &= \sup\{\|Ax\| : x\in K^n \mbox{ with }\|x\|= 1\} \\ &= \sup\left\{\frac{\|Ax\|}{\|x\|} : x\in K^n \mbox{ with }x\ne 0\right\}. \end{align}

Can someone provide a reference to the proof of this claim?

Fraïssé
  • 11,656

2 Answers2

2

You can also prove it by relating spectral radius with matrix norm. For a linear operator $A$, $$\operatorname{spr}(A) \le \|A\|$$ where $$\begin{align} \|A\| &= \underset{\|x\|=1}{\sup} \|Ax\| \\ \operatorname{spr}(A) &= \max\left\lbrace |\lambda|: \lambda \text{ is an eigenvalue of } A \right\rbrace \end{align}$$ The spectral radius $\operatorname{spr}(A)$ and $\|A\|$ are equal iff the operator $A$ is normal.

So $\|A\| < 1 \Rightarrow \operatorname{spr}(A) < 1 \Rightarrow $ all of the eigenvalues of $A$ have absolute value smaller than 1.

See Exercise I.2.6 of Rajendra Bhatia's Matrix Analysis where he calls it the Neumann Series. enter image description here

seeker
  • 297
1

See comments above: a proof is given on page 348 of Matrix Analysis by Horn and Johnson.

Citation: Horn, Roger A., and Charles R. Johnson. Matrix Analysis. 2nd ed. New York: Cambridge University Press, 2013, p 348

Ben Grossmann
  • 234,171
  • 12
  • 184
  • 355