(i) uma base e a dimensão de U.
(ii) uma base e a dimensão de W.
(iii) uma base e a dimensão de U+W.
(iv) uma base e a dimensão de U cap W.
(a) U=(x, y, z) in R^3 / x+y+z=0, W=(x, y, 0) / x, y in R, V=R^3.
(b) U=(x, y, z) in R^4 / x+y=t-z=0, W=(x, y, z, t) in R^4 / z=t=0, V=R^4.
(c) U=(x, y, z) in R^3 / x=0, W=[(2,2,0),(1,2,3),(7,12,21),(-1,-2,-3)], V=R^3.
(d) U=(x, y, z, t) in R^4 / x-y+t+z=0, W=(x, y, z, t) in R^4 / x+y-t+z=0, V=R^4.

Passo 1

O enunciado pede, para cada caso ((a)–(d)), que determinemos as subespaços U e W de V, suas bases e dimensões, depois a soma U+W e a interseção Ucap W também com base e dimensão. Vamos usar parametrizações de sistemas lineares para achar bases de U e W, resolver o sistema conjunto para Ucap W, e a fórmula
dim(U+W)=dim U+dim W-dim(Ucap W)
para obter dim(U+W) e então fornecer uma base, muitas vezes escolhendo a base padrão de R^n quando U+W=V.

Passo 2

Caso (a): V= R^3.
  • Para U=(x,y,z)mid x+y+z=0 resolvemos z=-x-y, logo
(x,y,z)=x(1,0,-1)+y(0,1,-1),
base (1,0,-1),(0,1,-1), dim U=2.
  • Para W=(x,y,0), claramente
(x,y,0)=x(1,0,0)+y(0,1,0),
base (1,0,0),(0,1,0), dim W=2.
  • Interseção Ucap W: z=0 e x+y+0=0implies y=-x, vetor genérico (x,-x,0)=x(1,-1,0), base (1,-1,0), dim(Ucap W)=1.
  • Soma U+W: dim(U+W)=2+2-1=3, logo U+W= R^3. Podemos tomar a base padrão (1,0,0),(0,1,0),(0,0,1).

Passo 3

Caso (b): V= R^4.
  • Para U=(x,y,z,t)mid x+y=0,;t-z=0, temos y=-x,;t=z,
(x,y,z,t)=x(1,-1,0,0)+z(0,0,1,1),
base (1,-1,0,0),(0,0,1,1), dim U=2.
  • Para W=(x,y,z,t)mid z=0,;t=0,
(x,y,0,0)=x(1,0,0,0)+y(0,1,0,0),
base (1,0,0,0),(0,1,0,0), dim W=2.
  • Interseção: z=t=0,;y=-x (x,-x,0,0)=x(1,-1,0,0), base (1,-1,0,0), dim(Ucap W)=1.
  • Soma: dim(U+W)=2+2-1=3. Uma base é por exemplo (1,0,0,0),(0,1,0,0),(0,0,1,1).

Passo 4

Caso (c): V= R^3.
  • U=(x,y,z)mid x=0 Rightarrow (0,y,z)=y(0,1,0)+z(0,0,1), base (0,1,0),(0,0,1), dim U=2.
  • W=span(2,2,0),(1,2,3),(7,12,21),(-1,-2,-3). Reduzindo, (-1,-2,-3)=-1cdot(1,2,3) e det[(2,2,0),(1,2,3),(7,12,21)]=12neq0, então base (2,2,0),(1,2,3),(7,12,21), dim W=3.
  • Como dim W=3 em R^3, W= R^3. Logo Ucap W=U, base (0,1,0),(0,0,1), dim(Ucap W)=2.
  • Soma U+W= R^3, base padrão (1,0,0),(0,1,0),(0,0,1), dim(U+W)=3.

Passo 5

Caso (d):
base , .
  • . De segue
base , .
  • Interseção: resolvendo o sistema
subtraindo dá . Logo
base , .
  • Soma: , mas , logo . Base padrão , .

Resposta

(a)(i) Base de U: (1,0,-1),(0,1,-1), dim U=2.
(a)(ii) Base de W: (1,0,0),(0,1,0), dim W=2.
(a)(iii) Base de U+W: (1,0,0),(0,1,0),(0,0,1), dim(U+W)=3.
(a)(iv) Base de Ucap W: (1,-1,0), dim(Ucap W)=1.
(b)(i) Base de U: (1,-1,0,0),(0,0,1,1), dim U=2.
(b)(ii) Base de W: (1,0,0,0),(0,1,0,0), dim W=2.
(b)(iii) Base de U+W: (1,0,0,0),(0,1,0,0),(0,0,1,1), dim(U+W)=3.
(b)(iv) Base de Ucap W: (1,-1,0,0), dim(Ucap W)=1.
(c)(i) Base de U: (0,1,0),(0,0,1), dim U=2.
(c)(ii) Base de W: (2,2,0),(1,2,3),(7,12,21), dim W=3.
(c)(iii) Base de U+W: (1,0,0),(0,1,0),(0,0,1), dim(U+W)=3.
(c)(iv) Base de Ucap W: (0,1,0),(0,0,1), dim(Ucap W)=2.
(d)(i) Base de U: (1,1,0,0),(-1,0,1,0),(-1,0,0,1), dim U=3.
(d)(ii) Base de W: (-1,1,0,0),(-1,0,1,0),(1,0,0,1), dim W=3.
(d)(iii) Base de U+W: (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1), dim(U+W)=4.
(d)(iv) Base de Ucap W: (-1,0,1,0),(0,1,0,1), dim(Ucap W)=2.

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