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## 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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