Is it always true that if $I+A$ is Non singular then $I-A$ is also Non singular Is it always true that if $I+A$ is Non singular then $I-A$ is also Non singular?
Since $I+A$ is Non singular we have
$$(I+A)^{-1}=I-A+A^2-A^3+A^4-A^5+\cdots$$ that is
$$(I+A)^{-1}=(I-A)+A^2(I-A)+A^4(I-A)+\cdots$$ $\implies$
$$(I+A)^{-1}=(I-A)(I+A^2+A^4+\cdots)$$
   taking Determinant on both sides we get
$$\frac{1}{Det(I+A)}=Det(I-A)Det(I+A^2+A^4+\cdots)$$  
since LHS is non zero, RHS should be non zero which means $Det(I-A)$ cannot be zero. hence $I-A$ is also Non singular. Is this proof valid?
 A: If you use divergent series, you can prove astonishing results. For instance, if
$$
S=1+2+4+\dots+2^n+\dotsb
$$
then
$$
2S=2+4+8+\dots+2^{n+1}+\dotsb=S-1
$$
whence $S=-1$. This is obviously nonsense (in the real numbers) and the problem is exactly in the series not being convergent.
Your matrix series $I-A+A^2-A^3+\dotsb$ isn't convergent either, unless all eigenvalues of $A$ have modulus less than $1$. Of course, in this case $1$ cannot be an eigenvalue of $A$, so $I-A$ is nonsingular.
Note that $I+I$ is nonsingular, but obviously $I-I$ is singular.
A: No, that fails if for example $A=I$. What happens is that the series does not converge.
However if the series $\sum A^j$ convergse absolutely we know that $(I-A)\sum{A^j} = I = (I+A)\sum{(-A)^j} = I$. This happens if the norm of $A$ is less than $1$. If the norm is larger than $1$ the series will definitely diverge (but that does not necessarily mean that $I+A$ is singular, just that if it's non-singular that the inverse can't be found using the series).
So what you can say is that if $|A|<1$ then both $I+A$ and $I-A$ are non-singular.
For $|A|=1$ then $I+A$ and $I-A$ may or may not be non-singular. It is for example non-singular if $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ then both $I+A$ and $I-A$ are non-singular. Tho case where $A-I$ is singuar is of course $A=I$.
Note: the norm of a matrix is the same as the modulus of the largest eigenvalue (if it has eigenvalues).
