Solution to Linear Algebra Done Wrong Exercise 1.3.5

Let $A$ be a linear transformation. If ${\bf z}$ is the center of the straight interval $[{\bf x}, {\bf y}]$, show that $A{\bf z}$ is the center of the interval $[A{\bf x}, A{\bf y}]$. Hint: What does it mean that ${\bf z}$ is the center of the interval $[{\bf x}, {\bf y}]$?

Solution: Note that ${\bf z}$ is the center of the interval $[{\bf x}, {\bf y}]$ if and only if $${\bf z}=\frac{{\bf x}+{\bf y}}{2}.$$Since $A$ is a linear transformation, we have \begin{align*}A{\bf z}=&\ A\left(\frac{{\bf x}+{\bf y}}{2}\right)\\ =& \ \frac{1}{2}(A{\bf x}+A{\bf y})\\ =&\ \frac{1}{2}\left(A{\bf x}+A{\bf y}\right).\end{align*}Hence $A{\bf z}$ is the center of the interval $[A{\bf x}, A{\bf y}]$.