True or False: If $A + A^2$ is invertible, then $A$ is also invertible A is a square $n$ by $n$ matrix here. 
I understand the proof for $A^2$ being invertible given that $A$ is invertible, but I fail to see how to incorporate the $A + A^2$ factor into it. 
What I have tried so far is a rough factoring to give:
$A(I_n + A)$
But that is where I am stuck. Any help is much appreciated!
 A: Yes. Note that $A+A^2=A(I+A)$. For the Binet rule:
$0\neq \det(A+A^2)=\det(A)\det(I+A)\Longrightarrow \det(A)\neq 0$.
A: If $A+A^2$ is invertible exist $B$ such that $(A+A^2)B=I\to A(I+A)B=I$ and then $A^{-1}=(I+A)B$ ie, $A$ is invertible.
A: Recall that a matrix is invertible if and only if it has a trivial nullspace.  Suppose $A+A^2$ is invertible.  We will prove that the nullspace of $A$ is $\{0\}$.
Suppose $Ax = 0$.  Then, $Ax+A(Ax) = 0 + A0 = 0$.  But, $Ax+A(Ax) = (A+A^2)x$, and since $A+A^2$ was assumed to be invertible, it must be that $x = 0$.  Thus, the nullspace of $A$ is $\{0\}$, so $A$ is invertible.
A: This will be true for every ring: if $a+a^2$ is invertible then so is $a$. It follows from another general fact: if $ab$ has a right inverse and $ca$ has a left inverse then $a$ is invertible. Indeed, $(a b) b_1=1$ implies $a(b b_1) = 1$ and $c_1 (ca) = 1$ implies $(c_1 c) a= 1$, so $a$ has both left and right inverses and so they coincide ( standard proof) and $a$ is invertible.
Now, to the proof of the statement. If $a+a^2$ is invertible then $a(1+a)$ is invertible and $(1+a)a$ is invertible, and by the above it follows that $a$ itself is invertible.
A: Yes.
This is a direct consequence of:

Proposition. Let $M$ denote a Dedekind-finite monoid. Then:
  
  
*
  
*For all $a,b \in M$, if $ab$ is invertible, then so too are $a$ and $b$.
  
*In other words, the invertible elements of $M$ form a saturated submonoid of $M$.
  

Proof. Suppose $ab$ is invertible. We will show that $a$ is invertible. Since $ab$ is invertible, We can find $k$ with $(ab)k=1$. So $a(bk)=1$. So by Dedekind-finiteness, we have that $a$ is invertible.
Returning to your problem, we use the fact that for all natural numbers $n$, the matrix ring $\mathbb{R}^{n \times n}$ is Dedekind-finite. Now, we know that
$$A+A^2 = A(I+A)$$
and it is given that the LHS is invertible. Hence by the above proposition, both $A$ and $I+A$ are invertible.
