1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Subspace proof in field F4

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex]b \in F[/itex] then [itex]{(x_1, x_2, x_3, x_4)\in F_4 : x_3 = 5x_4 + b}[/itex] is a subspace of [itex]F_4[/itex] if and only if b = 0.

    F is defined as either ℝ or ℂ.

    3. The attempt at a solution

    I'm still trying to get the hang of these "proofs."

    Let [itex]c \in F [/itex] to check if this is closed under scalar multiplication:

    [itex]c(x_1, x_2, x_3, x_4) = (c(x_1),c(x_2),c(x_3),c(x_4))[/itex]

    Then let [itex] y = (y_1,y_2,y_3,y_4) \in F_4 [/itex] and [itex]x = (x_1, x_2, x_3, x_4)[/itex] to check closure under addition:

    [itex] x + y = (x_1 + y_1, x_2 + y_2, x_3+y_3,x_4+y_4) [/itex]

    But since [itex]x_3 = 5x_4 + b[/itex] then we can substitute in the previous calculation where b will never get added to anything element in [itex] F_4 [/itex] because [itex] b \in F.[/itex] Doing this would increase the dimension(?) of [itex] F_4 [/itex] to 5 and thus not be in a subspace of [itex] F_4. [/itex] So, b = 0 for this to stay closed under addition in [itex] F_4 .[/itex]

    Is that correct? Thanks for any help.
    Last edited by a moderator: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2


    Staff: Mentor

    "ℝ or ℂ" renders as squares in my browser. What are these symbols?
  4. Nov 22, 2011 #3
    It's R and C for real and complex fields.
  5. Nov 22, 2011 #4


    User Avatar
    Homework Helper

    not quite, for scalar multiplication just subtitute in the property directly

    so you have
    [tex] c(x_1, x_2, x_3, x_4) = (cx_1,cx_2,cx_3,cx_4) [/tex]

    now sub in x_3 = 5x_4+b
    [tex] c(x_1, x_2, 5x_4+b, x_4) = (cx_1,cx_2,c(5x_4+b
    ),cx_4) = (cx_1,cx_2,c5x_4+cb
    ,cx_4) [/tex]

    for what values of b is this vector in the subspace?

    you will also need to have a think about what it means exactly to prove "if and only if"
  6. Nov 22, 2011 #5
    Thanks for the replies.

    The value of b can only be 0 because it would be adding another element to the list of [tex] F_4[/tex]
  7. Nov 22, 2011 #6


    User Avatar
    Homework Helper


    so you need to show
    F4 a subspace implies b=0
    b= 0 implies F4 is a subspace

    However in this case, just going through each of of the subspace requirements should be sufficient
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook