扁頭科學 發問時間: 科學數學 · 9 年前

線性代數特論(about subspace)

Let V be a vector space over F and let U be a subspace of V.

Define v+U = { v+u | u∈U }, where v∈V.

(1) Show that v+U is a subspace of V <==> v∈U.

(2) For v, w∈V, show that the following are equivalent:

(a) (v+U) ∩ (w+U) ≠ Ø (b) v-w ∈U (c) v+U = w+U.

2 個解答

評分
  • Sam
    Lv 6
    9 年前
    最佳解答

    Let V be a vector space over F and let U be a subspace ofV.

    Define v+U = { v+u | u∈U }, where v∈V.

    (1) Show that v+U is a subspace of V <==> v∈U.

    Lemma. obviously, if u∈U, then u+U=U.

    [pf of (1)](part 1 = >)Since v+Uis a subspace, 0∈v+U.= > thereexist a u in U, such that 0=v+u,= > v=-uin U.[[done ofpart 1]](part 2 < = )Since v inU, v+U=U is a subspace of V.[[done ofpart 2]][[DONE of pfof (1)]] (2) For v, w∈V, show that the following are equivalent:

    (a) (v+U) ∩ (w+U) ≠ Ø (b) v-w ∈U (c) v+U = w+U.[Pf of (2)](part 1 (a)= > (b) )From (a), letx∈ (v+U) ∩ (w+U).there areu1, u2 ∈ U, such thatx=v+u1=w+u2, = > v-w=u2-u1 ∈ U.[[done of part 1]] [part 2 (b)= > (c)]From (b),thereexist a u∈U, such that v-w=u, or v=w+u.v+U=(w+u)+U=w+(u+U)=w+U[[done of part 2]] [part 3 (c)= > (a)]From (c), v+Uㄈ w+U.Since 0∈U, v=v+0 ∈v+Uㄈ w+U.= > v∈w+U,= > v∈ (v+U)∩ (w+U)= > (v+U) ∩ (w+U) ≠ Ø[[done of part 3]] [[DONE of pfof (2)]]

  • 9 年前

    請你去想想代數學中,我們怎麼證factor group的性質

    這個題目很像!!

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