Quick Proof Verification: Showing Matrix is Nonsingular.

Question

Hello everyone I just want to verify if my proof is correct as I'm very worried that it's not: Thanks so much.

Question: Let $A$ be a square matrix and let $\lambda$ be a complex number for which $|\lambda| >\|A\|$, where $\|\cdot \|$ denotes some induced matrix norm. Show that the matrix $\lambda I-A$ is nonsingular and that

$({\lambda I-A})^{-1}= \frac{1}{\lambda}\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k$

Proof: Since $\|A\| \geq 0$ and $|\lambda| >\|A\|$ implies $\lambda$ implies $\lambda\neq 0$ $\|\frac{1}{\lambda}A\|<1$

$\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k$ converges. Call $C=\frac{1}{\lambda}\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k$

Now: $({\lambda I-A})C={\lambda}C-AC =\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k -\sum_{k=0}^{\infty}\frac{1}{\lambda^{k+1}}A^{k+1}=I$.

Obviously I would also have to show that $C({\lambda I-A})=I$ as well, but it reduces to the same two series subtracted from each other. The problem I'm having is understanding why the two series subtracted above=Identity. I know that that they should., I'm just not sure why exactly. Any help would be much appreciated. Thank you.

Answer 1

\begin{align} &\sum_{k=0}^\infty \frac{1}{\lambda^k}A^k - \sum_{k=0}^\infty \frac{1}{\lambda^{k+1}}A^{k+1}\\ &= \left(I+ \sum_{k=1}^\infty \frac{1}{\lambda^k}A^k\right) - \sum_{l=1}^\infty \frac{1}{\lambda^{l}}A^{l}\\ &=I+\left( \sum_{k=1}^\infty \frac{1}{\lambda^k}A^k- \sum_{l=1}^\infty \frac{1}{\lambda^{l}}A^{l}\right)\\ \\&=I \end{align}

where I have let $l=k+1$.

Answer 2

$\sum_{k=0}^{\infty}B^k-\sum_{k=0}^{\infty}B^{k+1}=$

$=I+B+B^2+B^3 +...-(B+B^2+B^3+...)=I$.

Answer 3

$$\sum_{k=0}^\infty\lambda^{-(k+1)}A^{k+1} =\sum_{k=1}^\infty\lambda^{-k}A^k=-I+\sum_{k=0}^\infty \lambda^{-k}A^k$$ etc. Or \begin{align} \sum_{k=0}^\infty \lambda^{-k}A^k-\sum_{k=0}^\infty \lambda^{-(k+1)}A^{k+1}&= I+\lambda^{-1}A+\lambda^{-2}A^2+\lambda^{-3}A^3+\cdots\\ &{ }-(\lambda^{-1}A+\lambda^{-2}A^2+\lambda^{-3}A^3+\cdots)\\ &=I. \end{align}