 jelentetvq

2022-01-24

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

Expert

span $\left(\left\{\stackrel{\to }{{v}_{1}},\stackrel{\to }{{v}_{2}}\right\}\right)=\left\{\lambda \stackrel{\to }{{v}_{1}}\mid \lambda \in F\right\}$
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 $\left\{\stackrel{\to }{{v}_{1}},\stackrel{\to }{{v}_{2}}\right\}$ 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
$span\left(S\right)=\left\{\sum _{i=1}^{k}{\lambda }_{k}{s}_{k}\mid n\in \mathbb{N},{s}_{i}\in S,{\lambda }_{i}\in F\right\}$
(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:
$span\left(\left\{\stackrel{\to }{{v}_{1}},\stackrel{\to }{{v}_{2}}\right\}\right)=\left\{\sum _{i=1}^{k}{\lambda }_{i}{v}_{i}\mid {\lambda }_{i}\in F\right\}$
$=\left\{{\lambda }_{1}\stackrel{\to }{{v}_{1}}+{\lambda }_{2}\stackrel{\to }{{v}_{2}}\mid {\lambda }_{1},{\lambda }_{2}\in F\right\}$
But note that $\stackrel{\to }{{v}_{2}}=2\stackrel{\to }{{v}_{1}}$, and so, for any ${\lambda }_{1},{\lambda }_{2}\in F$
${\lambda }_{1}\stackrel{\to }{{v}_{1}}+{\lambda }_{2}\stackrel{\to }{{v}_{2}}={\lambda }_{1}\stackrel{\to }{{v}_{1}}+{\lambda }_{2}\left(2\stackrel{\to }{{v}_{1}}\right)=\left({\lambda }_{1}+2{\lambda }_{2}\right)\stackrel{\to }{{v}_{1}}$
Then, as any linear combination of $\left\{\stackrel{\to }{{v}_{1}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{\to }{{v}_{2}}\right\}$ can be expressed as a scalar multiple of $\left\{\stackrel{\to }{{v}_{1}}$, and any scalar multiple of $\left\{\stackrel{\to }{{v}_{1}}$ can be expressed as a linear combination of $\left\{\stackrel{\to }{{v}_{1}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{\to }{{v}_{2}}\right\}$ by setting ${\lambda }_{2}=0$ we have
$\left(\left\{\stackrel{\to }{{v}_{1}},\stackrel{\to }{{v}_{2}}\right\}\right)=\left\{\lambda \stackrel{\to }{{v}_{1}}\mid \lambda \in F\right\}$

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