# show that {$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$:$x_3=5x_4+b$} is a subspace

Let $b\in\mathbb{F}$ then {$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$:$x_3=5x_4+b$} is a subspace iff $b=0$. This is from axler's Linear algebra done right. $\mathbb{F}$ stands for a field either the real numbers or complex numbers. I proved the reverse direction but now I have to show the foward direction. This is what I have so far.

($\rightarrow$) Assume that $b\in\mathbb{F}$ and let T={$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$: $x_3=5x_4+b$} where $T$ is a subspace of $\mathbb{F}^4$. Suppose not that $b\neq 0$. Since $T$ is a subspace by assumption that means that $0\in T$ That is $(x_1,x_2,x_3,x_4)=(0,0,0,0)$ This means that $x_3=0$ and $x_4=0$. If $x_3=0$ then we have $0=5x_3+b \implies \frac{5}{b}=x_4$ but since $b\neq 0$ we have that $x_4\neq 0$. Suppose that $x_4=0$. Again we have $x_3=5(0)+b=b$ Since $b\neq 0$ this means $x_3\neq 0$. Thus a contradiction. Would this be right?

What you've got is sufficient, but it could be cleaned up a bit. If $b \neq 0$, then $x_3 = x_4 = 0$ is a contradiction, so $T$ does not contain the zero vector and is not a subspace.