Answer
(a) ${u} \cdot {v}={u}^{T} {v}=-6$.
(b) ${v} \cdot {v}={v}^{T} {v}=8$.
(c) $\|u\|^{2}=5$.
(d) $({u} \cdot {v})v=\left[\begin{array}{l}{-12}\\ {12}\end{array}\right]$.
(e) ${u} \cdot {(5v)}=-30$.
Work Step by Step
Given ${u}=\left[\begin{array}{l}{-1} \\ {2}\end{array}\right]$ and ${v}=\left[\begin{array}{l}{2}\\ {-2}\end{array}\right]$, then we have
(a) ${u} \cdot {v}={u}^{T} {v}=[-1 \ \ 2]\left[\begin{array}{l}{2} \\ {-2}\end{array}\right]=[(-1)(2)+(2)(-2)]=-6$.
(b) ${v} \cdot {v}={v}^{T} {v}=[2 \ \ -2]\left[\begin{array}{l}{2} \\ {-2}\end{array}\right]=[(2)(2)+(-2)(-2)]=8$.
(c) $\|u\|^{2}={u} \cdot {u}={u}^{T} {u}=[-1 \ \ 2]\left[\begin{array}{l}{-1} \\ {2}\end{array}\right]=[(-1)(-1)+(2)(2)]=5$.
(d) $({u} \cdot {v})v=({u}^{T} {v}) v=-6\left[\begin{array}{l}{2}\\ {-2}\end{array}\right]=\left[\begin{array}{l}{-12}\\ {12}\end{array}\right]$.
(e) ${u} \cdot {(5v)}={u}^{T} {(5v)}=[-1 \ \ 2]\left[\begin{array}{l}{10} \\ {-10}\end{array}\right]=[(-1)(10)+(2)(-10)]=-30$.
