Use o processo de Gram-Shmidt para transformar a base V1, V2, V3, V4 numa base ortonormal.
V1=(0,2,1,0), V2=(1,-1,0,0), V3=(1,2,0,-1), V4=(1,0,0,1)

Passo 1

O enunciado dá a base
V1=(0,2,1,0),;V2=(1,-1,0,0),;V3=(1,2,0,-1),;V4=(1,0,0,1)
em R^4 com o produto interno usual, e pede uma base ortonormal via Gram–Schmidt. Vamos usar a projeção
proju(v)=lang<= v,urang<=lang<= u,urang<=,u
e depois normalizar cada vetor.

Passo 2

Definimos
u1=V1=(0,2,1,0),
|u1|=raiz0^2+2^2+1^2+0^2=raiz5, logo
e1=u1|u1|=1raiz5(0,2,1,0)

Passo 3

Calculamos
lang<= V2,u1rang<==1cdot0+(-1)cdot2+0cdot1+0cdot0=-2,
lang<= u1,u1rang<==5,
u2=V2-lang<= V2,u1rang<=lang<= u1,u1rang<=,u1 =(1,-1,0,0)-l(-tfrac25r)(0,2,1,0) =(1,-tfrac15,tfrac25,0)
Para facilitar, multiplicamos por 5 e obtemos (5,-1,2,0) com norma raiz30. Então
e2=1raiz30(5,-1,2,0)

Passo 4

Calculamos
lang<= V3,u1rang<==1cdot0+2cdot2+0cdot1+(-1)cdot0=4,
lang<= V3,u2rang<==1cdot5+2cdot(-1)+0cdot2+(-1)cdot0=3.
Logo
u3 =V3-45u1-330u2 =(1,2,0,-1)-45(0,2,1,0)-110(5,-1,2,0) =(tfrac12,tfrac12,-1,-1)
Multiplicando por 2, fica (1,1,-2,-2) com norma raiz10. Portanto
e3=1raiz10(1,1,-2,-2)

Passo 5

Finalmente
lang<= V4,u1rang<==0,
lang<= V4,u2rang<==1cdot5+0cdot(-1)+0cdot2+1cdot0=5,
lang<= V4,u3rang<==1cdot1+0cdot1+0cdot(-2)+1cdot(-2)=-1.
Então
u4 =V4-530u2--110u3 =(1,0,0,1)-tfrac16(5,-1,2,0)+tfrac110(1,1,-2,-2) =(tfrac415,tfrac415,-tfrac815,tfrac45)
Multiplicando por 15 e simplificando por 4, fica (1,1,-2,3) com norma raiz15. Logo
e4=1raiz15(1,1,-2,3)

Resposta

Uma base ortonormal é
e1=1raiz5(0,2,1,0), e2=1raiz30(5,-1,2,0), e3=1raiz10(1,1,-2,-2), e4=1raiz15(1,1,-2,3)

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