Prove $I - A$, $I + A$ and $I - A^2$ are nonsingular I need to prove $I - A$, $I + A$ and $I - A^2$  are nonsingular where $||A|| < 1$.
Also need to show $(I - A)^{-1} = \sum_{k=0}^{\infty}A^k$.
So far I've got that because $||A||< 1$ then $p(A) < 1$. Therefore $p(I-A) < 1$ (as $p(I) = 1$). I want this lead to me stating that this means none of the eigenvalues of $I-A$ are $0$, therefore $A$ is nonsingular.
For $(I - A)^{-1} = \sum_{k=0}^{\infty}A^k$:
If I say $(I-A)\sum_{k=0}^{m}A^k = I - A^{m+1}$
Then $\sum_{k=0}^{m}A^k = (I - A)^{-1}(I - A^{m+1})$ and because $||A|| < 1$ then $I-A^{m+1}$ tends to $0$ as $m$ tends to inf.
I appreciate my answers are far from complete and appreciate any help!
 A: Computing an explicit inverse by a norm convergent geometric power series is legitimate but you need to adapt your argument a little: at one point you are confusing the $m$-th partial sum of a series with the infinite sum.
The following argument rather follows the "eigenvalue" line that you indicated.
If one of these three matrices is singular then that matrix has a nontrivial kernel, i.e., at least one eigenvector with eigenvalue $0.$ By linearity this means that $A,$ or $-A,$ or $A^2$ has an eigenvector with eigenvalue $\pm1$ and this means that the norm of that matrix cannot be less than $1.$ Also remember that the norm of $A^2$ is bounded by the square of the norm of $A.$
A: Show the second point first. For all $n \in \mathbb{N}$ and all matrices with $\|B\| < 1$, we have 
$$(I-B)\sum_{k=0}^{n} B^{k} = \sum_{k=0}^{n} B^{k} (I-B) =  I - B^{n+1}$$
Because $\|B\| < 1$, taking the limit yields 
$$(I-B)^{-1} = \sum_{k=0}^{\infty}B^{k}$$
which shows that $(I-B)$ is invertible. Now apply this reasoning to $B = A, B= -A, B= A^2$.
A: Suppose $(A-I)x = 0$. Without loss of generality we can take $\|x\|=1$. Then $Ax = x$, which implies $\|A\| \ge 1$ a contradiction. Hence $\ker (A-I) $ is trivial and so $A-I$ is invertible.
Exactly the same argument applies to $A+I$.
Since $I-A^2 = (I-A)(I+A)$ we see that $I-A^2$ is also invertible since it is the product of two invertible matrices.
For the second part we can use Gelfand's formula which says that
$\lim_{k} \sqrt[k]{\|A^k\|} = \rho(A)$.
In particular, if $\rho(A)<\lambda <1$, then there is some $K$ such that
for all $k \ge K$ we have $\|A^k\| \le \lambda^k$,
and so $\lim_k A^k = 0$.
