Proving $A$ is invertible if $A + A^2 = I$. I'm trying to prove $A$ is invertible by proving there is an $A'$, for $AA' = I$.
So I got to this stage $A(I + A) = I$, now I determine that $A' = I + A$, 
and from that I get $AA' = I$.
I wanted to know if this is valid,  casue it doesn't seem to make sense,and I can't find a 'real' solution for such equation.
Note:  $A$ is $n \times n$ and  $I$ represents the identity matrix and is also of the same dimensions.
 A: We have $1=\det{(I)}=\det{(A+A^2)}=\det{(A(I+A))}=\det({A)}\det{(I+A)}$. Thus $\det({A)}\neq 0$ and $A$ is invertible.
A: Given: $A + A^2 = I$.
You set out to prove $A$ is invertible by proving there is an $A'$, such that $AA' = I$.  
You recognized that $A + A^2 = AI + A^2 = A(I +A) = I.$
So you got to this stage: $A(I + A) = I$. Now you let $A^{\prime} = I + A$, and from that you got $AA' = I$, where $A^{\prime}$ is the right inverse of $A$.
Note that, because addition of matrices is commutative, $$A+A^2 = A^2 + A = (A+I)A = = (I + A)A=I.$$ So by the same strategy we used to show that $(I + A)$ is a right inverse of $A$, it follows that $(I+A)$ is also a left inverse of $A$, and hence is THE unique inverse of $A.$
In general, and as you may already know:

For any square matrix $M$, if $M$ has a right inverse, then the right-inverse is the unique inverse of $M$, and so it is also a left inverse of $M$ (and vice versa).

So you did indeed find the inverse of $A$, in having found a right inverse of $A$, namely $A^{\prime} = I + A = A^{-1}$. 

Therefore, since $A$ has an inverse, $A$ is invertible.

An alternate strategy is to take the determinant of each side of the equation:
$A + A^2 = I.$
Note that $$1 = \det{(I)} = \det{(A + A^2)} = \det{(A(I + A))} = \det{(A)}\det{(I + A)} > 0.$$
So $\det{(A)}$ cannot be zero. Thus, $A$ is invertible.

What might be confusing you is that you are proving IF $A + A^2 = I$, THEN $A$ is invertible, but that is NOT to say that if $A$ is invertible, then it is always the case $A + A^2 = I$.
So you shouldn't expect to find that $A+A^2 = I$ for all invertible matrices $A$, and don't worry if an example of an invertible matrix $A$ for which $A+A^2 = I$ doesn't immediately come to mind. 

A: From $A+A^2=I$ you do indeed get that $AA'=I$ for some $A'$, namely $A'=I+A$. And since the distribution laws hold "on both sides", you also get $A'A=I$ for the same $A'$, which proves that $A$ is invertible.
A: With $A(I+A)=I$ you’re almost there: this shows that $A$ has a right inverse, $I+A$. But a square matrix has a right inverse if and only if it’s actually invertible, so $A$ is invertible.
If you don’t already know that theorem, you can prove it in a variety of ways. For instance, $$I=I^T=\big(A(I+A)\big)^T=(I+A)^TA^T\;,$$ so $A^T$ has a left inverse. Suppose that $A^Tx=0$. Then $$x=Ix=(I+A)^TA^Tx=(I+A)^T0=0\;,$$ so the null space of $A^T$ is trivial, containing only the zero vector. Therefore $A^T$ is invertible, and its inverse must be $(I+A)^T$. Thus, $I=A^T(I+A)^T=\big((I+A)A\big)^T$, and therefore $$(I+A)A=I^T=I\;,$$ meaning that $I+A$ is also a left inverse of $A$. Since $I+A$ is both a left and a right inverse of $A$, it’s actually $A^{-1}$, and $A$ is invertible.
Added: As tst points out in the comments, I was working much too hard here. We can factor the $A$ out of $A+A^2$ on the right as well as on the left, so not only do we have $A(I+A)=I$, we also have $(I+A)A=I$, and it’s immediate that $I+A=A^{-1}$.
A: Put $f(x) = x^2 + x - 1$.  We have $f(A) = 0$.  Since $x = 0$ is not a root of $f$, zero is not an eigenvalue of $A$.  Thus, $A$ is invertible.  
A: Hint $\ $ Over any ring: polynomial $\rm\,f(a) = 0\:$ and $\rm\:f(0)\:$ invertible $\rm\:\Rightarrow\: a\:$ invertible (with two-sided inverse, if $\rm\:a\:$ commutes with all coefficients $\rm\,c_i\,$ of $\rm\,f),\,$ since
$$\rm\:c_n a^n + \cdots + c_1 a + c_0 = 0\,\ \Rightarrow\,\ (c_n a^{n-1}+\cdots + c_1) a\, =\, -c_0$$
thus left-multiplying by $\rm\:-c_0^{-1}$ yields a left-inverse for $\rm\:a.\:$ When $\rm\:a\:$ commutes with the coefficients, we can commute $\rm\:a\:$ to the left, showing that the above inverse is a two-sided inverse (in particular such commutativity is true if the coefficients are integers, so universally commutative).
A: Prove if $A+A^2 = I$ then $A$ is invertible
Contrapositive
If $A$ is singular then $A+A^2 \neq I$
we have $detA = 0$ and $det|A+A^2| = det|A|det|A+I| = 0$
And $detI = 1$
So $A+A^2 \neq I$
