Orthogonality

1 min read Updated Wed May 06 2026 19:36:36 GMT+0000 (Coordinated Universal Time)

Ortho is a Greek prefix meaning right.

Definition

For 2 vectors

When their inner product is 00. Angle between the vectors are 9090^\circ.

For a set of vectors

When all pairs of vectors are orthogonal.

An orthogonal set is automatically linearly-independent.

Orthonormal

When 2 vectors or all pairs of vectors of a set are:

  • orthogonal and
  • norm of all vectors is 1

Gram–Schmidt Orthogonalization

A method to convert a linearly independent set into a orthogonal set.

Suppose {x1,,xn}\set{x_1,\dots,x_n} is linearly independent.

yk=xkj<kxk,yjyj2yj.y_k = x_k - \sum_{j<k} \frac{\langle x_k, y_j\rangle}{|y_j|^2} y_j.

For p,q=11p(x)q(x)dx\langle p,q\rangle = \int_{-1}^{1} p(x)q(x)\,\text{d}x, its orthogonal set is named Legendre polynomials of the first kind.

{pn(x)}={1,x,12(3x21),}\set{ p_n(x) } = \bigg\{ 1, x, \frac{1}{2}(3x^2-1), \dots \bigg\}

nn starts from 0. Additionally pn(x)p_n(x) also satisfies:

(1x2)y2xy+x(x+1)y=0(1-x^2)y'' - 2xy' + x(x+1)y = 0

For p,q=0exp(x)q(x)dx\langle p,q\rangle = \int_{0}^{\infty} e^{-x}p(x)q(x)\,\text{d}x, its orthogonal set is named Leguerre polynomials of the first kind.

{Ln(x)}={1,1x,12(x24x+2)}\set{ L_n(x) } = \set { 1, 1-x, \frac{1}{2}(x^2-4x+2) }

nn starts from 0. Additionally Ln(x)L_n(x) satisfies:

xy(1x)y+xy=0xy'' - (1-x)y' + xy = 0