Question

Step 2. Now we show that $T$ is closed under vector addition. Suppose that $v_1$ and $v_2$ belong to $T$. We want to show that $v_1 + v_2$ also belongs to $T$. To show this, firstly note that we know these vectors must be of the form $v_1 = s \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$ and $v_2 = t \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$ for some $s, t \in \mathbb{R}$. So we have $v_1 + v_2 = s \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix} + t \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix} = \mu \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$ where (in terms of $s$ and $t$) $\mu = 3v_1 - v_2 + 2v_3 \in \mathbb{R}$. Thus $v_1 + v_2$ is of the form required, and so we have shown $T$ is closed under vector addition.

          Step 2. Now we show that $T$ is closed under vector addition. Suppose that $v_1$ and $v_2$ belong to
$T$. We want to show that $v_1 + v_2$ also belongs to $T$. To show this, firstly note that we know
these vectors must be of the form
$v_1 = s \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$ and $v_2 = t \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$
for some $s, t \in \mathbb{R}$. So we have
$v_1 + v_2 = s \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix} + t \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix} = \mu \begin{pmatrix} 3 \ -1 \ 2 \end{pmatrix}$
where (in terms of $s$ and $t$)
$\mu = 3*v_1 - v_2 + 2*v_3 \in \mathbb{R}$.
Thus $v_1 + v_2$ is of the form required, and so we have shown $T$ is closed under vector addition.

Show more…

Step 2. Now we show that T is closed under vector addition. Suppose that v1 and v2 belong to
T. We want to show that v1 + v2 also belongs to T. To show this, firstly note that we know
these vectors must be of the form
v1 = s
< p m a t r i x > and v2 = t
< p m a t r i x >
for some s, t ∈ℝ. So we have
v1 + v2 = s
< p m a t r i x >
+ t
< p m a t r i x >
= μ
< p m a t r i x >
where (in terms of s and t)
μ = 3*v1 - v2 + 2*v3 ∈ℝ.
Thus v1 + v2 is of the form required, and so we have shown T is closed under vector addition.

Added by Miguel M.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Ben Blakesley

01/25/2022

Step 1

This means that $\mathbf{v}_1$ and $\mathbf{v}_2$ can be written in the form $s \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ and $t \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ respectively, where $s$ and $t$ are real numbers. Show more…

Show all steps

lock

Thanks for your feedback!

Step 2. Now we show that $T$ is closed under vector addition. Suppose that $\mathbf{v}_1$ and $\mathbf{v}_2$ belong to $T$ . We want to show that $\mathbf{v}_1 + \mathbf{v}_2$ also belongs to $T$. To show this, firstly note that we know these vectors must be of the form $\mathbf{v}_1 = s \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{v}_2 = t \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ for some $s, t \in \mathbb{R}$. So we have $\mathbf{v}_1 + \mathbf{v}_2 = s \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \mu \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ where (in terms of $s$ and $t$) $\mu = 3\mathbf{v}_1 - \mathbf{v}_2 + 2\mathbf{v}_3$ ∈ R. Thus $\mathbf{v}_1 + \mathbf{v}_2$ is of the form required, and so we have shown $T$ is closed under vector addition.

Play audio

Feedback

Powered by NumerAI

Ben Blakesley and 63 other subject Algebra educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later