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**Glitch** **The question:**

Show that the set

$\displaystyle S = \{x \in \mathbb{R}^4 | x = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$ for some $\displaystyle \lambda \in \mathbb{R}\}$

is a subspace of $\displaystyle \mathbb{R}$

**My attempt:**

Using Subspace Theorem:

S clearly contains the zero vector in $\displaystyle \mathbb{R}^4$ since the set is all solutions to the parametric form of a line through the origin.

Let **x**,**y** $\displaystyle \in$ S

**x** + **y** = $\displaystyle \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$

= $\displaystyle 2\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$

Since $\displaystyle \lambda \in \mathbb{R}$ and $\displaystyle 2 \in \mathbb{R}$, $\displaystyle 2\lambda \in \mathbb{R}$, so

= $\displaystyle \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$

(I'm unsure of that reasoning)

Therefore closed under vector addition.

Let $\displaystyle \alpha \in \mathbb{R}, x \in S$

$\displaystyle \alphax = \alpha(\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix})$

= $\displaystyle \alpha\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$

= Since $\displaystyle \lambda \in \mathbb{R}$ and $\displaystyle \alpha \in \mathbb{R}, \alpha\lambda \in \mathbb{R}$, so

= $\displaystyle \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$

Therefore closed under scalar multiplication.

Is this proof sound? Thanks.