jelentetvq

Answered

2022-01-24

Let $\overrightarrow{{v}_{1}}=\left[\begin{array}{c}2\\ 3\end{array}\right]$ and $\overrightarrow{{v}_{1}}=\left[\begin{array}{c}4\\ 6\end{array}\right]$ what is the **span** of the vector space defined by $\overrightarrow{{v}_{1}}$ and $\overrightarrow{{v}_{1}}$ ? Explain your answer in detail?

Answer & Explanation

Gordon Stephens

Expert

2022-01-25Added 10 answers

span $\left(\{\overrightarrow{{v}_{1}},\overrightarrow{{v}_{2}}\}\right)=\{\lambda \overrightarrow{{v}_{1}}\mid \lambda \in F\}$

Explanation: Typically we talk about the span of a set of vectors, rather than of an entire vector space. We will proceed, then, in examining the span of$\{\overrightarrow{{v}_{1}},\overrightarrow{{v}_{2}}\}$ within a given vector space.

The span of a set of vectors in a vector space is the set of all finite linear combinations of those vectors. That is, given a subset S of a vector space over a field F, we have

$span\left(S\right)=\{\sum _{i=1}^{k}{\lambda}_{k}{s}_{k}\mid n\in \mathbb{N},{s}_{i}\in S,{\lambda}_{i}\in F\}$

(the set of any finite sum with each term being the product of a scalar and an element of S)

For simplicity, we will assume that our given vector space is over some subfield F of$\mathbb{C}$ . Then, applying the above definition:

$span\left(\{\overrightarrow{{v}_{1}},\overrightarrow{{v}_{2}}\}\right)=\{\sum _{i=1}^{k}{\lambda}_{i}{v}_{i}\mid {\lambda}_{i}\in F\}$

$=\{{\lambda}_{1}\overrightarrow{{v}_{1}}+{\lambda}_{2}\overrightarrow{{v}_{2}}\mid {\lambda}_{1},{\lambda}_{2}\in F\}$

But note that$\overrightarrow{{v}_{2}}=2\overrightarrow{{v}_{1}}$ , and so, for any ${\lambda}_{1},{\lambda}_{2}\in F$

$\lambda}_{1}\overrightarrow{{v}_{1}}+{\lambda}_{2}\overrightarrow{{v}_{2}}={\lambda}_{1}\overrightarrow{{v}_{1}}+{\lambda}_{2}\left(2\overrightarrow{{v}_{1}}\right)=({\lambda}_{1}+2{\lambda}_{2})\overrightarrow{{v}_{1}$

Then, as any linear combination of$\left\{\overrightarrow{{v}_{1}}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\overrightarrow{{v}_{2}}\right\}$ can be expressed as a scalar multiple of $\{\overrightarrow{{v}_{1}}$ , and any scalar multiple of $\{\overrightarrow{{v}_{1}}$ can be expressed as a linear combination of $\left\{\overrightarrow{{v}_{1}}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\overrightarrow{{v}_{2}}\right\}$ by setting ${\lambda}_{2}=0$ we have

$\left(\{\overrightarrow{{v}_{1}},\overrightarrow{{v}_{2}}\}\right)=\{\lambda \overrightarrow{{v}_{1}}\mid \lambda \in F\}$

Explanation: Typically we talk about the span of a set of vectors, rather than of an entire vector space. We will proceed, then, in examining the span of

The span of a set of vectors in a vector space is the set of all finite linear combinations of those vectors. That is, given a subset S of a vector space over a field F, we have

(the set of any finite sum with each term being the product of a scalar and an element of S)

For simplicity, we will assume that our given vector space is over some subfield F of

But note that

Then, as any linear combination of

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