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Let $A$ be a complex $n \times n $ matrix. Let $\lambda_1, \cdots, \lambda_n$ be the eigenvalues of $A$. If $|z|> \max_{k=1}^n|\lambda_k|$, express $(zI-A)^{-1}$ as a convergent power series.

Usually we have if $|r<1|$, then $\frac{1}{1-r}=\sum_{n=1}^{\infty} r^n$.

If we do the same thing, we get $(zI-A)^{-1} = \frac{1}{I-A/z}$. However, I'm not sure if $A/z$ makes sense here. Also, I think we need to use the fact the that eigenvalues are bounded.

Any help will be appreciated!

Korn
  • 1,558
  • Since $|z|>\rho(A)$, we have $\rho(\frac{1}{z}A)<1$. Therefore the Neumann series $\sum_{n=0}^\infty\left(\frac{1}{z}A\right)^n$ converges to $(I-\frac{1}{z}A)^{-1}$. Hence $(zI-A)^{-1}=\frac{1}{z}(I-\frac{1}{z}A)^{-1}$ can be expanded as a Laurent series in $z$. Note that this is not a power series in $z$ because there is a $\frac{1}{z}$ term, but it is a power series in $A$. – user1551 Aug 08 '22 at 02:55
  • https://math.stackexchange.com/q/2077772/27978 – copper.hat Aug 08 '22 at 04:25

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