Prove that if $\rho(A)<1 \implies \frac 1{I_n - A} = \sum_{k = 0}^{\infty}A^k$

Question

Prove that if $\rho(A)<1 \implies \frac 1{I_n - A} = \sum_{k = 0}^{\infty}A^k$.

My attempt:

Well if $\rho(A) < 1\implies |\lambda_{max}|<1$, but $A$ can be rewritten as: $$A=PDP^{-1}$$

Where $D$ is a diagonal matrix with the corresponding eigenvalues. Then

$$A^k=PD^kP^{-1}$$

But if $|\lambda_{max}|<1$ then $|\lambda(A)|<1 \implies |\lambda(A)|^k<1$ so:

$$\lim_{k\to\infty}D^k=O_n$$

Then the series must converge like that.

Answer 1

Yes, your attempt is correct partially.

Missing argument:

Partial sums are $I+A+A^2+...+A^n$, now we see that, $$(I-A)(I+A+A^2+...+A^n)=I-A^{n+1}$$ But $A^{n+1}\longrightarrow O_n$, $$\Longrightarrow (I-A)(I+A+A^2+...)=I$$ Hence the result.

Hope it helps:)