Let $U$ be an $m \times n$ matrix where the columns of $U$ form an orthonormal set

Question

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.

Answer 1

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.

Answer 2

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): $$\langle Ux,Uy\rangle = \sum_{ijk} U_{ij}x_j U_{ik}y_k= \sum_{jk} x_j \delta_{j,k} y_k = \sum_j x_j y_j = \langle x,y\rangle.$$