Suppose $A$ and $B$ are square matrices and that $AB=A$ with $B \neq I$. What does this say about the invertibility of $A$?
This question showed up on an exam I took this past spring. I got stuck on it, but I thought about it for a while and think I figured it out. Here's something similar to what I got:
Suppose $A$ is invertible. Then:
$$\begin{align} AB &= A \\ A^{-1}AB &= A^{-1}A \\ IB &= I \\ B &= I \end{align}$$
This shows that if $AB=A$, then $B$ must be an identity matrix if $A$ is invertible.
Conclusion: If $AB=A$ and $B \neq I$, then $A$ must be singular.
An obvious example would be making $A$ a zero matrix.
Is what I've got correct?
What you write is right. However, one should not forget that matrix calculus is only a computational aspect of linear algebra, which itself is closely related to elementary geometry. Another example than the one you propose (the null matrix) comes spontaneously to my mind: the projections in a vector space $E$ can be defined as the linear applications $p:E \to E$ such that $p^2=p$. Consider the elementary case of the projection in $E=\mathbb{R}^2$, "the plane", defined as follows: let {i,j} be the canonical basis of $E$. $p$ is the projection on $\mathbb{R}i$ parallel to $\mathbb{R}j$. Then the matrix of p in the canonical basis is $P=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$. And we have $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ So with $A=P$ and $B=P$, $$AB=A, B\neq I, A \text{ singular.}$$