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Subspace proof in field F4

  • #1
658
2

Homework Statement


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.

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.
 
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Answers and Replies

  • #2
33,309
5,001

Homework Statement


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.
"ℝ or ℂ" renders as squares in my browser. What are these symbols?

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.
 
  • #3
658
2
It's R and C for real and complex fields.
 
  • #4
lanedance
Homework Helper
3,304
2

Homework Statement


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.

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.
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"
 
  • #5
658
2
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]
 
  • #6
lanedance
Homework Helper
3,304
2
yep

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
 

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