Can F4 Be a Subspace If b Is Nonzero?

[DrummingAtom]

Homework Statement

Show that if $b \in F$ then ${(x_1, x_2, x_3, x_4)\in F_4 : x_3 = 5x_4 + b}$ is a subspace of $F_4$ 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 $c \in F$ to check if this is closed under scalar multiplication:

$c(x_1, x_2, x_3, x_4) = (c(x_1),c(x_2),c(x_3),c(x_4))$

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

$x + y = (x_1 + y_1, x_2 + y_2, x_3+y_3,x_4+y_4)$

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

Is that correct? Thanks for any help.

 

[Mark44]

Homework Statement

Show that if $b \in F$ then ${(x_1, x_2, x_3, x_4)\in F_4 : x_3 = 5x_4 + b}$ is a subspace of $F_4$ 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 $c \in F$ to check if this is closed under scalar multiplication:

$c(x_1, x_2, x_3, x_4) = (c(x_1),c(x_2),c(x_3),c(x_4))$

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

$x + y = (x_1 + y_1, x_2 + y_2, x_3+y_3,x_4+y_4)$

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

Is that correct? Thanks for any help.

 

[DrummingAtom]

It's R and C for real and complex fields.

 

[lanedance]

Homework Statement

Show that if $b \in F$ then ${(x_1, x_2, x_3, x_4)\in F_4 : x_3 = 5x_4 + b}$ is a subspace of $F_4$ 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 $c \in F$ to check if this is closed under scalar multiplication:

$c(x_1, x_2, x_3, x_4) = (c(x_1),c(x_2),c(x_3),c(x_4))$

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

$x + y = (x_1 + y_1, x_2 + y_2, x_3+y_3,x_4+y_4)$

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

Is that correct? Thanks for any help.

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

so you have
$$c(x_1, x_2, x_3, x_4) = (cx_1,cx_2,cx_3,cx_4)$$

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

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"

 

[DrummingAtom]

Thanks for the replies.

The value of b can only be 0 because it would be adding another element to the list of $$F_4$$

 

[lanedance]

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