Orthogonality

Ortho is a Greek prefix meaning right.

Definition

For 2 Vectors

When their inner product is 00. Angle between the vectors are 90∘90^\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=xk−∑j<k⟨xk,yj⟩∣yj∣2yj.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(3x2−1),…}\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:

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

For ⟨p,q⟩=∫0∞e−xp(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) }={ 1,1−x,12(x2−4x+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′′−(1−x)y′+xy=0xy'' - (1-x)y' + xy = 0
Written by September 29, 2026 1 min read
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