Problem 14 Show that if $S_{1}$ and \(S_{... [FREE SOLUTION]

Chapter 1: Problem 14

Show that if $S_{1}$ and $S_{2}$ are arbitrary subsets of a vector space V, then $\operatorname{span}\left(S_{1} \cup S_{2}\right)=\operatorname{span}\left(S_{1}\right)+\operatorname{span}\left(S_{2}\right)$. (The sum of two subsets is defined in the exercises of Section 1.3.)

Short Answer

Expert verified

To show that $\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)$, we first showed that $\operatorname{span}(S_1 \cup S_2)$ is a subset of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ by considering an arbitrary vector $v$ in $\operatorname{span}(S_1 \cup S_2)$ and showing it can be written as a sum of vectors in $\operatorname{span}(S_1)$ and $\operatorname{span}(S_2)$. Next, we showed that $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ is a subset of $\operatorname{span}(S_1 \cup S_2)$ by considering an arbitrary vector $v$ in $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ and showing it can be written as a linear combination of vectors in $S_1 \cup S_2$. Thus, $\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)$.

Step by step solution

Show that $\operatorname{span}(S_1 \cup S_2)$ is a subset of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$

First, let's consider an arbitrary vector $v$ in $\operatorname{span}(S_1 \cup S_2)$. This means that $v$ can be written as a linear combination of the vectors in $S_1 \cup S_2$. However, we can also write $v$ as a linear combination of the vectors in $S_1$ and $S_2$ separately. Now let's write $v$ as a sum of two vectors: \[v = v_1 + v_2\] where $v_1$ is a linear combination of vectors in $S_1$ and $v_2$ is a linear combination of vectors in $S_2$. Therefore, we can say: \[v_1 \in \operatorname{span}(S_1) \text{ and } v_2 \in \operatorname{span}(S_2)\] Since $v_1 \in \operatorname{span}(S_1)$ and $v_2 \in \operatorname{span}(S_2)$, we can conclude: \[v \in \operatorname{span}(S_1) + \operatorname{span}(S_2)\] Thus, if a vector is in $\operatorname{span}(S_1 \cup S_2)$ it must also be in $\operatorname{span}(S_1) + \operatorname{span}(S_2)$. This implies that $\operatorname{span}(S_1 \cup S_2)$ is a subset of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$. Now let's show that $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ is a subset of $\operatorname{span}(S_1 \cup S_2)$.

Show that $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ is a subset of $\operatorname{span}(S_1 \cup S_2)$

Let's consider an arbitrary vector $v$ in $\operatorname{span}(S_1) + \operatorname{span}(S_2)$. By definition, $v$ can be written as a sum of two vectors: \[v = v_1 + v_2\] where $v_1 \in \operatorname{span}(S_1)$ and $v_2 \in \operatorname{span}(S_2)$. Now, since $v_1 \in \operatorname{span}(S_1)$, $v_1$ should be a linear combination of the vectors in $S_1$. Similarly, since $v_2 \in \operatorname{span}(S_2)$, $v_2$ should be a linear combination of the vectors in $S_2$. Thus, $v$ should be a linear combination of the vectors in $S_1 \cup S_2$, which implies: \[v \in \operatorname{span}(S_1 \cup S_2)\] Thus, if a vector is in $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ it must also be in $\operatorname{span}(S_1 \cup S_2)$. This implies that $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ is a subset of $\operatorname{span}(S_1 \cup S_2)$.

Conclusion

Since we have shown that $\operatorname{span}(S_1 \cup S_2)$ is a subset of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ and that $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ is a subset of $\operatorname{span}(S_1 \cup S_2)$, it follows that these two sets are actually equal: \[\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)\] This concludes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Show that if \(S_{1}\) and \(S_{2}\) are arbitrary subsets of a vector space V, then $\operatorname{span}\left(S_{1} \cup S_{2}\right)=\operatorname{span}\left(S_{1}\right)+\operatorname{span}\left(S_{2}\right)$. (The sum of two subsets is defined in the exercises of Section 1.3.)

Short Answer

Step by step solution

Show that \(\operatorname{span}(S_1 \cup S_2)\) is a subset of \(\operatorname{span}(S_1) + \operatorname{span}(S_2)\)

Show that \(\operatorname{span}(S_1) + \operatorname{span}(S_2)\) is a subset of \(\operatorname{span}(S_1 \cup S_2)\)

Conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Pure Maths

Probability and Statistics

Discrete Mathematics

Applied Mathematics

Statistics

Study anywhere. Anytime. Across all devices.