# Subspace proof

• December 14th 2010, 09:09 PM
Glitch
Subspace proof
The question:
Show that the set
$S = \{x \in \mathbb{R}^4 | x = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$ for some $\lambda \in \mathbb{R}\}$
is a subspace of $\mathbb{R}$

My attempt:
Using Subspace Theorem:

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

Let x,y $\in$ S
x + y = $\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
= $2\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
Since $\lambda \in \mathbb{R}$ and $2 \in \mathbb{R}$, $2\lambda \in \mathbb{R}$, so
= $\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
(I'm unsure of that reasoning)

Let $\alpha \in \mathbb{R}, x \in S$
$\alphax = \alpha(\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix})$
= $\alpha\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
= Since $\lambda \in \mathbb{R}$ and $\alpha \in \mathbb{R}, \alpha\lambda \in \mathbb{R}$, so
= $\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
Therefore closed under scalar multiplication.

Is this proof sound? Thanks.
• December 14th 2010, 09:12 PM
Drexel28
Quote:

Originally Posted by Glitch
The question:
Show that the set
$S = \{x \in \mathbb{R}^4 | x = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$ for some $\lambda \in \mathbb{R}\}$
is a subspace of $\mathbb{R}$

My attempt:
Using Subspace Theorem:

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

Let x,y $\in$ S
x + y = $\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
= $2\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
Since $\lambda \in \mathbb{R}$ and $2 \in \mathbb{R}$, $2\lambda \in \mathbb{R}$, so
= $\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
(I'm unsure of that reasoning)

Let $\alpha \in \mathbb{R}, x \in S$
$\alphax = \alpha(\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix})$
= $\alpha\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
= Since $\lambda \in \mathbb{R}$ and $\alpha \in \mathbb{R}, \alpha\lambda \in \mathbb{R}$, so
= $\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
Therefore closed under scalar multiplication.

Is this proof sound? Thanks.

I think you have a fundamental misunderstanding of the notation. The $\lambda$ is not fixed. So, for example to show that two elements of your set are closed you need to show that $\lambda_1 M+\lambda_2 M=\lambda_3 M$ where $M$ is the vector in the problem.
• December 14th 2010, 09:24 PM
Glitch
Ahh, I see. So for vector addition, it should be:

x + y = $\lambda_1M + \lambda_2M$
= $(\lambda_1 + \lambda_2)M$
Since $(\lambda_1 + \lambda_2) \in R$, S is closed under vector addition.

Correct?
• December 14th 2010, 09:25 PM
Drexel28
Quote:

Originally Posted by Glitch
Ahh, I see. So for vector addition, it should be:

x + y = $\lambda_1M + \lambda_2M$
= $(\lambda_1 + \lambda_2)M$
Since $(\lambda_1 + \lambda_2) \in R$, S is closed under vector addition.

Correct?

Exactly!
• December 14th 2010, 09:26 PM
Glitch
Thanks. I knew I was going something silly!