Series of Matrices question Let $A\in L(\mathbb{R}^n\rightarrow \mathbb{R}^n) $ and $ \left \| A \right \|_\mathrm{op}<1$.
$$S=(\operatorname{Id}-A+A^2-A^3+A^4-\cdots)$$

Prove that $S$ converges and $S(\operatorname{Id}+A)=(\operatorname{Id}+A)S=\operatorname{Id}$. Also prove that $\det(A+Id)>0$.

So I got that S converges using the triangle inequality so $S < \operatorname{Id} + \sum \left \| A \right \|^n<\infty$
The second part is where I'm stuck now..
I know that if I'll prove that I will get that $\operatorname{Id}+A=S^{-1}$ and is obviously invertible so $\det(\operatorname{Id}+A)$ can't be $0$, not sure how to prove its positive.
 A: Define 
$$
S_m = \sum_{k=0}^m (-1)^k A^k.
$$
Then $S = \lim_{m\to\infty} S_m$, and, for example, $S(I + A) = I$ iff $S_m(I+A) = I$ as $m \to \infty$. Can you get it from here?
AMENDMENT:
Since $S(I+A) = I$ from above, taking determinants and using $\det(AB) = (\det A)(\det B)$, we have $(\det S) \det(I + A) = 1$. From this we conclude $\det(I+A) > 0$ iff $\det S > 0$. We'll show $\det S > 0$. 
The determinant of a linear operator on an $n$-dimensional vector space is the product of its $n$ eigenvalues (with multiplicity): if $\Lambda(S)$ denotes the multiset of $n$ (not necessarily distinct) eigenvalues of $S$, then
$$
\det S = \prod_{\mu \in \Lambda(S)} \mu.
$$
So we have to show that the product of the eigenvalues of $S$ is positive. Note that the $n$ eigenvalues $\mu \in \Lambda(S)$ of $S$ come from the $n$ eigenvalues $\lambda \in \Lambda(A)$ of $A$:
\begin{align*}
A v = \lambda v &\implies (-A)^k v = (-\lambda)^k v \\
&\implies \sum_{k=0}^\infty (-A)^k v = \sum_{k=0}^\infty (-\lambda)^k v \\
&\implies Sv = \underbrace{\frac{1}{1+\lambda}}_{\mu = (1 + \lambda)^{-1}} v.
\end{align*}
In this calculation, we have used the fact that 
$$
\|A\|_\text{op} < 1 \implies |\lambda| < 1
$$
[can you show this?] to conclude that the geometric series converges. 
Hence
$$
\det S = \prod_{\mu \in \Lambda(S)} \mu = \prod_{\lambda \in \Lambda(A)} \frac{1}{1+\lambda}.
$$
We break into cases $\lambda \in \mathbb{R}$ and $\lambda \in \mathbb{C}\backslash\mathbb{R}$.
If $\lambda \in \mathbb R$, then again using $|\lambda| < 1$, we note that 
$$
\frac{1}{1+\lambda} > 0,
$$
contributing a positive factor to $\det S$. Good.
If $\lambda \in \mathbb C \backslash \mathbb R$ is an eigenvalue of $A$, then so is its complex conjugate $\bar \lambda$ because $A$ acts on a real vector space. This pair contributes a factor of
$$
\frac{1}{1 + \lambda} \cdot \frac{1}{1+ \bar \lambda} 
$$
to $\det S$. This is positive iff the denominator is. The denominator may be written as
\begin{align*}
(1 + \lambda)(1 + \bar \lambda)
&= (1 + \lambda) (\overline{1 + \lambda}) \\
&= \left| 1 + \lambda \right|^2,
\end{align*}
so it is indeed positive, completing the proof that $\det S > 0$. 
