Orthogonal sets - Orthogonality and Least Squares

Notes:

A set of vectors {$v_1,\cdots,v_n$} in $\Bbb{R}^n$ are orthogonal sets if each pair of vectors from the set are orthogonal. In other words,
$v_i \cdot v_j =0$ Where $i \neq j$.

If the set of vectors {$v_1,\cdots,v_n$} in $\Bbb{R}^n$ is an orthogonal set, then the vectors are linearly independent. Thus, the vectors form a basis for a subspace $S$. We call this the orthogonal basis.

To check if a set is an orthogonal basis in $\Bbb{R}^n$, simply verify if it is an orthogonal set.
$y=c_1 v_1+c_2 v_2+\cdots+c_p v_p$
Are calculated by using the formula:
$c_i = \frac{y \cdot v_i}{v_i \cdot v_i}$ where $i=1,\cdots,p$.

A set {$v_1,\cdots,v_p$}is an orthonormal set if it's an orthogonal set of unit vectors.

If $S$ is a subspace spanned by this set, then we say that {$v_1,\cdots,v_p$} is an orthonormal basis. This is because each of the vectors are already linear independent.

A $m \times n$ matrix $U$ has orthonormal columns if and only if $U^T U=I$.

Let $U$ be an $m \times n$ matrix with orthonormal columns, and let $x$ and $y$ be in $\Bbb{R}^n$. Then the 3 following things are true:

1) $\lVert Ux \rVert = \lVert x \rVert$

2) $(Ux) \cdot (Uy)=x \cdot y$

3) $(Ux) \cdot (Uy)=0$ if and only if $x \cdot y =0$

Consider $L$ to be the subspace spanned by the vector $v$ . Then the orthogonal projection of $y$ onto $v$ is calculated to be:
$\hat{y}=$proj$_Ly=\frac{y \cdot v}{v \cdot v}v$
The component of $y$ orthogonal to $v$ (denoted as $z$) would be:
$z=y-\hat{y}$