Orthogonal

Consider 2 column matrices $v_1$ and $v_2$:

$$v_1= \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} \; \land \; v_2= \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}$$

Product

The product of $v_1$ and $v_2$ is defined as:

$$v_1\cdot v_2=\sum_{k=1}^n{a_kb_k}=v_2\cdot v_1 = v_1^{T}v_2$$

Orthogonal vectors

$v_1$ and $v_2$ are orthogonal iff $v_1\cdot v_2 = 0$.

For a set of $n$ column vectors, they are orthogonal iff they are pairwise orthogonal. That is: $$

$$\forall i,j \in \set{1,\dots,n} \land i \neq j, (v_i\cdot v_j = 0)$$

Note

$v_1, v_2$ are orthogonal $\implies v_1, v_2$ are linearly independent.

Converse is not true.

Orthogonal matrix

For a square matrix $A$ with real entries, it is orthogonal iff $A^{-1}=A^{T}$.

A matrix is orthogonal iff sum of the squared elements of any row or column is $1$. $$

Properties

• $\det{A} = \pm 1$
• $A$ is invertible, non-singular
• $A^{-1} = A^{T}$
• $A^T, A^{-1}$ are orthogonal
• It is diagonalizable over $\mathbb{C}$ (may not be, over $\mathbb{R}$)
• $\text{rank}\,A=\text{order}\,A$
• Product of 2 orthogonal matrices of the same order is also an orthogonal matrix
• The columns or rows of an orthogonal matrix form an orthogonal set of vectors