Is $\{(x_1,x_2):x_1-5x_2=0\}$ a subspace of $\Bbb{R}^2$ ?
I know the zero vector is in the subspace because if $x_1 = 0$ and $x_2=0$ then it is in the subspace.
is it allowed to set $5x_1 - 5x_2 = 0$ which would prove the vector addition clause.
I guess i feel pretty lost and I'm trying to understand how to determine the subset to be able to answer the question
To show it is subspace, you have to show that the zero vector (in this case, the vector = (0,0)) is in $U = \{ (x_1,x_2) : x_1 - 5x_2 = 0 \} \subset \mathbb{R}^2$, that if you take two vectors ${\bf x} \in U$ and ${\bf x'} \in U$ then ${\bf x+x'} \in U$ and if $\alpha \in \mathbb{R}$ then $\alpha {\bf x} \in U$.
Clearly, $0-5\cdot 0 = 0$ thus the vector $(0,0) \in U$.
Now, take two vectors in $U$, say ${\bf x} = (x_1,x_2)$ and ${\bf x'} = (x_1',x_2') \in U$. This means that $x_1 - 5x_2 = 0$ and $x_1' - 5x_2' = 0$
Now, I want to show that ${\bf x + x'} \in U$. In other words, I want to show that $(x_1+x_1')-5(x_2+x_2')=0$. Indeed,
\begin{align*} (x_1+x_1')-5(x_2+x_2') &= x_1+x_1' - 5x_2-5x_2' \\ &= (x_1-5x_2) + (x_1'-5x_2') \\ &= 0 + 0\\ &= 0 \end{align*}
Now, if $\alpha$ is a scalar, and ${\bf x} \in U$ can you show that $\alpha {\bf x}$ is in $U$? That is, you want to show that $\alpha x_1 - 5 (\alpha x_2) = 0$. Isn't it obvious?