Mustn't both left and right inverses be checked?

Question

Source: Linear Algebra by Lay (4 edn 2011). p. 160. Chapter 2 Supplementary Exercise 4.

Exercise: Suppose $A^n = 0$ matrix for some $n > 1$. Find an inverse for $I - A$.

Solution: From p. 160 Supplementary Exercise 3, the inverse of $I-A$ is probably $I+A+A^{2}+...+A^{n-1}$. To verify this, compute $ (I \color{orangered}{-A} )\color{forestgreen}{(I+A+\cdots+A^{n-1})}=I+A+\cdots+A^{n-1} \quad \color{orangered}{-A}(I+A+\cdots+A^{n-1})=I \color{orangered}{-A} A^{n-1}=I-A^{n}. $
If $A^{n}=0$, then the above equation becomes $(I-A)\color{forestgreen}{\sum_{0 \le i \le n - 1} A^{i}} =...=I$. Since $I-A$ and $\sum_{0 \le i \le n - 1} A^{i} $ are square, they are invertible by the Invertible Matrix Theorem, and $\sum_{0 \le i \le n - 1} A^{i} $ is the inverse of $I-A$. $\blacksquare$

$1.$ The definition of inverse overhead (from Lay p. 103) contains two conjuncts. Why didn't the solution fail to check that $\color{forestgreen}{\sum_{0 \le i \le n - 1} A^{i}}$ is the left inverse? Why only check it as the right inverse?

$2.$ Why didn't Lay define with only one of the right or left inverse, because If $AB = I$ then $BA = I$ guarantees the other inverse? Is Lay's definition with two conjuncts/inverses redundant?

Answer 1

In principle, if $A$ is a square matrix then $AC=I\iff CA=I$. So, once you have left/right inverse for $A$, then you will have the other one.

Answer 2

You are correct that some care is needed.

For finite matrices, existence of left inverse implies invertibility, and existence of a right inverse implies invertibility. It is not very hard to show these claims but certainly if it was not established then you need to verify on both sides. In this particular case, the verification is essentially identical to the one given, and I guess that is why this detail was omitted. But you are correct, there is something else to say here.