Exercise 3.1.32

The columns of $\mathit{AB}$ are combinations of the columns of $A$. So all columns of $[\mathit{AAB}]$ are already in $C(A).$ But $A=\left[\begin{array}{cc}0\hfill & 1\hfill \\ 0\hfill & 0\hfill \end{array}\right]$has a larger column space than $A^2=\left[\begin{array}{cc}0\hfill & 0\hfill \\ 0\hfill & 0\hfill \end{array}\right]$. For square matrices, the column space is $R^n$ exactly when $A$ is invertible.