TITLE: True or False: If $A + A^2$ is invertible, then $A$ is also invertible
QUESTION [7 upvotes]: 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!

REPLY [2 votes]: 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.