1. ## Confusing subspace problem

The question:

Show that T = { $\sum_{i=1}^4 \lambda _i v_i : \lambda _i \in \mathbb{R}, 1 \le i \le 4$}, where $v_1, v_2, v_3, v_4$ are given fixed vectors in $\mathbb{R}^3$ is either a subspace or not a subspace of $\mathbb{R}^3$.

I'm not sure what they mean by "given fixed vectors" means, or how to use them in my reasoning. I'm also a little confused as to how I can check the Subspace Theorem when there's summation involved. Any assistance would be great.

2. Originally Posted by Glitch
The question:

Show that T = { $\sum_{i=1}^4 \lambda _i v_i : \lambda _i \in \mathbb{R}, 1 \le i \le 4$}, where $v_1, v_2, v_3, v_4$ are given fixed vectors in $\mathbb{R}^3$ is either a subspace or not a subspace of $\mathbb{R}^3$.

I'm not sure what they mean by "given fixed vectors" means, or how to use them in my reasoning. I'm also a little confused as to how I can check the Subspace Theorem when there's summation involved. Any assistance would be great.
So I want you to try this on your own (and report back if you haven any trouble). An example of what the question is asking you to figure out is given the four 'fixed' (not changing) vectors $\left\{\begin{bmatrix}1\\ 0\\ 3\\ 4\end{bmatrix},\begin{bmatrix}0\\ 3\\ 28\\ 7\end{bmatrix},\begin{bmatrix}\pi\\ 37\\ e\\ 2\end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}\right\}$ you can, for each quadruple of 'fixed' (given, not changin) real numbers $r_1,r_2,r_3,r_4$ you can make a new vector $v_{r_1,r_2,r_3,r_4}=r_1\begin{bmatrix}1\\ 0\\ 3\\ 4\end{bmatrix}+r_2\begin{bmatrix}0\\ 3\\ 28\\ 7\end{bmatrix}+r_3\begin{bmatrix}\pi\\ 37\\ e\\ 2\end{bmatrix}+r_4\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}r_1+r_3\pi\\ 3r_2+37r_3\\ 3r_1+28r_2+er_3\\ 4r_1+7r_2+2r_3+r_4\end{bmatrix}$. You can then consider the set $\left\{v_{r_1,r_2,r_3,r_4}:r_1,r_2,r_3,r_4\in\math bb{R}\right\}$. Is that set closed under linear combinations?

3. I think I see what you're saying. This is what I have:

i) Let $v_i \in \mathbb{R}^3$
Let $\lambda_i = 0$
$\lambda_i v_i = 0v_i = 0$
Therefore contains zero vector

ii) $\sum_{i=1}^4\lambda_i v_i = \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 + \lambda_4 v_4$
Let $\alpha \in \mathbb{R}$

$\alpha \sum_{i=1}^4\lambda_i v_i = \alpha(\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 + \lambda_4 v_4)$

$\alpha \sum_{i=1}^4\lambda_i v_i = \alpha\lambda_1 v_1 + \alpha\lambda_2 v_2 + \alpha\lambda_3 v_3 + \alpha\lambda_4 v_4$

= $\sum_{i=1}^4\alpha\lambda_i v_i$
= $\sum_{i=1}^4(\alpha\lambda_i)v_i$
Since $\alpha\lambda_i \in \mathbb{R}$, set T is closed under scalar multiplication

iii) Let a, b $\in \mathbb{R}^3$
$\sum_{i=1}^4\lambda_i a_i + \sum_{i=1}^4\lambda_i b_i$
= $\sum_{i=1}^4\lambda_i a_i + \lambda_i b_i$
= $\sum_{i=1}^4\lambda_i(a_i + b_i)$
Since $a_i + b_i \in \mathbb{R}^3$, set T is closed under vector addition.

Therefore, T is a subspace of $\mathbb{R}^3$

is this proof correct? Thanks.