Exercise 1.3.2

We choose $A=\left[\begin{array}{cc}\hfill -4\hfill & \hfill -8\hfill \\ \hfill 2\hfill & \hfill 4\hfill \end{array}\right]$, so $A^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$. We have $\text{rank}(A)=1$, $\text{rank}(A^2)=0\text{rank}(A)$. We also can prove that $A^{T}A=\left[\begin{array}{cc}\hfill 20\hfill & \hfill 40\hfill \\ \hfill 40\hfill & \hfill 80\hfill \end{array}\right]$, so we see that $\mathit{rank}(A^{T}A)=\mathit{rank}(A)$. You can find such $A$ by making assumptions on its columns and the columns of $A^2$.