Let $U$ be an $m \times n$ matrix where the columns of $U$ form an orthonormal set 
Let $U$ be an $m \times n$ matrix where the columns of $U$ form an orthonormal set.
a) If $\vec{x}$ and $\vec{y}$ are in $\mathbb{R}^n$, show that $(U\vec{x})*(U\vec{y}) = \vec{x} * \vec{y}$
b) Show that $\|U(\vec{x})\| = \|\vec{x}\|$

I'm very confused on how to do this exactly. I'm looking for a non-calculus approach since this is a Linear Algebra class.
 A: I will assume you are working with the field of real numbers $\mathbb{R}$.  Let $\{\vec{u}_1,\ldots,\vec{u}_n\}$ denote the columns of $U$, i.e.,
$$U=\begin{pmatrix}
\vec{u}_1 & \vec{u}_2 &\cdots & \vec{u}_n
\end{pmatrix}.$$
Write
$$\vec{x}=\begin{pmatrix}
x_1 \\\vdots \\x_n
\end{pmatrix},
\qquad \vec{y}=\begin{pmatrix}
y_1 \\ \vdots \\ y_n
\end{pmatrix}$$
Then we have
$$(U\vec{x})\cdot (U\vec{y})=\left(\sum_{k=1}^nx_k\vec{u}_k\right)\cdot\left(\sum_{k=1}^ny_k\vec{u}_k\right)=\sum_{k=1}^n\sum_{j=1}^nx_ky_j\vec{u}_k\cdot\vec{u}_j=\sum_{k=1}^nx_ky_k=\vec{x}\cdot\vec{y}.$$
(Minor adjustments needed here to account for complex scalars).  Using this, the answer to part (b) is simple.
A: In principle already said by Rodrigo, but writing it out (we need $m\geq n$):
$$\sum_{i=1}^m U_{ij} U_{ik} = \delta_{j,k}, \ \ \ 1\leq j,k\leq n.$$
And then (omitting limits on the sums): $$\langle Ux,Uy\rangle = \sum_{ijk} U_{ij}x_j U_{ik}y_k= \sum_{jk} x_j \delta_{j,k} y_k = \sum_j x_j y_j = \langle x,y\rangle.$$
