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Orthogonal and Orthonormal Basis, Orthogonal Matrix,Orthogonal Compliment

Orthogonal Vectors

Two vectors are orthogonal if the angle between them is 90 degree or the dot product is zero.

Eg: [ 1,0] and [0,1] are two orthogonal vectors

[1,1] and [-1,1] are two orthogonal vectors

Orthonormal vectors

Two vectors are orthonormal if they are orthogonal and also the length of each vector is 1.

Eg: [ 1,0] and [0,1] are two orthonormal vectors ( length 1 and dot product is zero)

[1,1] and [-1,1] are two orthogonal vectors but not orthonormal( length is not 1)

Definition

Let S = {v1, v2, ... , vk} be a set of vectors in Rn, then S is called an orthogonal if
vi . vj = 0
for all i not equal to j. An orthogonal set of vectors is called orthonormal if all vectors in S are unit vectors.

Theorem

Any orthogonal set of vectors S = {v1, v2, ... , vk} are linearly independent.

Proof

Let

c1v1 + ... + cnvk = 0

Since the vectors are orthogonal, we have

vi . vj = 0 (for i != j)

and when we dot both sides of the equation with vi all the terms drop out except the ith term.

c1v1 . vi + ... + cnvk . vi = 0 . vi

= ci

And we see that

ci = 0

Hence S is linearly independent.

Finding the coordinate of a vector with respect to a basis can be computationally difficult, usually including the inverse of a matrix.  However if the basis is orthonormal, then the computation is simple.  

Theorem

Let S = {v1, v2, ... , vn} be an orthonormal basis for Rn and v be a vector in Rn then the ith coordinate of v with respect to the basis S is

([v]S)i = v . vi

Proof

Since S is a basis, it spans Rn hence we can write

v = c1v1 + ... + cnvk

now just take the dot product with vi on both sides and note and use the fact that S is orthonormal to get

v . vi = ci
Example
Find the coordinates for the vector (5,10) with respect to the basis
S = {(3/5, 4/5), (-4/5, 3/5)}
Solution

We find

(5,10) . (3/5,4/5) = 11 
 (5,10) . (-4/5, 3/5) = 2

so that

[(5,10)]S = (11,2)

Orthogonal Matrix
A matrix which contains orthonormal vectors as its column is called orthogonal matrix.

For an orthogonal matrix
$A.A^T=A^TA=I$ which implies that $A^{-1}=A^T$

Transformations by orthogonal matrices are special because the length of a vector $x$ is not changed when transforming it using an orthogonal matrix.
Moreover the angle between two vectors is unchanged when transforming both of them using an orthogonal matrix.
This means that orthogonal matrices A with $A^T=A^{-1}$ preserve both angles and distances.
Eg: The rotation matrix

import numpy as np
B=np.array([[1/np.sqrt(2),1/np.sqrt(2)],[1/np.sqrt(2),-1/np.sqrt(2)]])
print("Orthogonal Matrix B")
print(B)
print("B Inverse")
print(np.linalg.inv(B))
print("B Transpose")
print(B.T)
print("B x B.T= I")
print(B.dot(B.T))

o/p
Orthogonal Matrix B 
[[ 0.70710678 0.70710678] 
 [ 0.70710678 -0.70710678]]
 B Inverse
 [[ 0.70710678 0.70710678] 
 [ 0.70710678 -0.70710678]] 
B Transpose 
[[ 0.70710678 0.70710678] 
 [ 0.70710678 -0.70710678]] 
B x B.T= I
 [[1. 0.] 
 [0. 1.]]

Orthogonal Compliment

Having defined orthogonality, lets now look at vector spaces that are orthogonal to each other.This play a major role in dimensionality reduction.Consider $D$ dimensional vector space $V$ and an $M$ dimensional subspace $U \subseteq V$. then its orthogonal compliment $U^\perp$ is a $(D-M)$ dimensional subspace of $V$ and contains all vectors in $V$ that are orthogonal  to every vector in $U$.Furthermore, $U \cap U^\perp={0}$ so that any vector $x \in V$ can be uniquely decomposed into

$x=\sum_{m=1}^{M} \lambda_mb_m +  \sum_{j=1}^{D-M} \psi_ib_j^{\perp}$ $\lambda_m,\psi_j \in \mathbb{R}$

where $(b_1,\ldots,b_M)$  is a basis of $U$ and $(b_1^\perp,\ldots,b_{D-M}^\perp)$ is a basis of $U^\perp$.
Therefore, the orthogonal compliment can also be used to describe a plane $U$(two dimensional subspace) in a three dimensional vector space.More specifically, the vector $w$ with $||w||=1$, which is orthogonal to the plane $U$, is the basis vector of $U^\perp$.All vectors that are orthogonal to $w$ must lie in the plane $U$.The vector $w$ is called the normal vector of $U$.
Generally orthogonal compliments can be used to describe hyperplane in n-dimensional vector and affine spaces.

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