Ortonormalidad

¿Qué aspecto tiene un par de vectores ortonormales en el espacio euclidiano bidimensional?

Sea u = (x ₁ , y ₁ ) y v = (x ₂ , y ₂ ). Consideremos las restricciones sobre x ₁ , x ₂ , y ₁ , y ₂ necesarias para que u y v formen un par ortonormal.

• De la restricción de ortogonalidad, u • v = 0.
• De la restricción de longitud unitaria en u , || u || = 1.
• De la restricción de longitud unitaria en v , || v || = 1.

Al expandir estos términos se obtienen 3 ecuaciones:

1. ${\displaystyle x_{1}x_{2}+y_{1}y_{2}=0\quad }$
2. ${\displaystyle {\sqrt {{x_{1}}^{2}+{y_{1}}^{2}}}=1}$
3. ${\displaystyle {\sqrt {{x_{2}}^{2}+{y_{2}}^{2}}}=1}$

Convertir de coordenadas cartesianas a polares y considerar la ecuación${\displaystyle (2)}$y ecuación${\displaystyle (3)}$inmediatamente da como resultado r ₁ = r ₂ = 1. En otras palabras, exigir que los vectores tengan longitud unitaria restringe los vectores a estar sobre el círculo unitario .

Después de la sustitución, la ecuación${\displaystyle (1)}$se convierte ${\displaystyle \cos \theta _ {1}\cos \theta _ {2}+\sin \theta _ {1}\sin \theta _ {2}=0}$. Reorganizando se obtiene${\displaystyle \tan \theta _{1}=-\cot \theta _{2}}$. Usando una identidad trigonométrica para convertir el término cotangente se obtiene

${\displaystyle \tan(\theta _{1})=\tan \left(\theta _{2}+{\tfrac {\pi }{2}}\right)}$

${\displaystyle \Rightarrow \theta _{1}=\theta _{2}+{\tfrac {\pi }{2}}}$

Es evidente que, en el plano, los vectores ortonormales son simplemente radios del círculo unitario cuya diferencia de ángulos es igual a 90°.

Los conjuntos ortonormales poseen ciertas propiedades muy atractivas que hacen que sea particularmente fácil trabajar con ellos.

• Teorema . Si { e₁ _, e₂ _, ... , eₙ _} es una lista ortonormal de vectores, entonces$${\displaystyle \forall {\textbf {a}}:=[a_{1},\cdots ,a_{n}];\ \|a_{1}{\textbf {e}}_{1}+a_{2}{\textbf {e}}_{2}+\cdots +a_{n}{\textbf {e}}_{n}\|^{2}=|a_{1}|^{2}+|a_{2}|^{2}+\cdots +|a_{n}|^{2}}$$
• Teorema . Toda lista ortonormal de vectores es linealmente independiente .

• Teorema de Gram-Schmidt . Si { v ₁ , v ₂ ,..., v _n } es una lista linealmente independiente de vectores en un espacio con producto interno${\displaystyle {\mathcal {V}}}$, entonces existe una lista ortonormal { e ₁ , e ₂ ,..., e _n } de vectores en${\displaystyle {\mathcal {V}}}$de tal manera que span ( e ₁ , e ₂ ,..., e _n ) = span ( v ₁ , v ₂ ,..., v _n ).

Proof of the Gram-Schmidt theorem is constructive, and discussed at length elsewhere. The Gram-Schmidt theorem, together with the axiom of choice, guarantees that every vector space admits an orthonormal basis. This is possibly the most significant use of orthonormality, as this fact permits operators on inner-product spaces to be discussed in terms of their action on the space's orthonormal basis vectors. What results is a deep relationship between the diagonalizability of an operator and how it acts on the orthonormal basis vectors. This relationship is characterized by the Spectral Theorem.

The standard basis for the coordinate spaceFⁿ is

Any two vectors e_i, e_j where i≠j are orthogonal, and all vectors are clearly of unit length. So {e₁, e₂,...,e_n} forms an orthonormal basis.

When referring to real-valued functions, usually the L² inner product is assumed unless otherwise stated. Two functions ${\displaystyle \phi (x)}$ and ${\displaystyle \psi (x)}$ are orthonormal over the interval${\displaystyle [a,b]}$ if

${\displaystyle (1)\quad \langle \phi (x),\psi (x)\rangle =\int _{a}^{b}\phi (x)\psi (x)dx=0,\quad {\rm {y}}}$

${\displaystyle (2)\quad ||\phi (x)||_{2}=||\psi (x)||_{2}=\left[\int _{a}^{b}|\phi (x)|^{2}dx\right]^{\frac {1}{2}}=\left[\int _{a}^{b}|\psi (x)|^{2}dx\right]^{\frac {1}{2}}=1.}$

The Fourier series is a method of expressing a periodic function in terms of sinusoidal basis functions. Taking C[−π,π] to be the space of all real-valued functions continuous on the interval [−π,π] and taking the inner product to be

${\displaystyle \langle f,g\rangle =\int _{-\pi }^{\pi }f(x)g(x)dx}$

it can be shown that

${\displaystyle \left\{{\frac {1}{\sqrt {2\pi }}},{\frac {\sin(x)}{\sqrt {\pi }}},{\frac {\sin(2x)}{\sqrt {\pi }}},\ldots ,{\frac {\sin(nx)}{\sqrt {\pi }}},{\frac {\cos(x)}{\sqrt {\pi }}},{\frac {\cos(2x)}{\sqrt {\pi }}},\ldots ,{\frac {\cos(nx)}{\sqrt {\pi }}}\right\},\quad n\in \mathbb {N} }$

forms an orthonormal set.

However, this is of little consequence, because C[−π,π] is infinite-dimensional, and a finite set of vectors cannot span it. But, removing the restriction that n be finite makes the set dense in C[−π,π] and therefore an orthonormal basis of C[−π,π].