Question

If$\left(v_1,v_2,....v_n\right)$spans K over F and w is any element of K , show that role="math" localid="1656921214077" $\mathbf{(}\mathit{w}\mathbf{,}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$also spans K.

Short answer

Answer:

$\mathbf{(}\mathit{w}\mathbf{,}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$is spans K

Step-by-step solution

Definition of linear span.

The linear span also called as just span of a set of vector in a vector space is the intersection of all subspace containing the set.

Step 2:Showing that (w,v1,v2,....vn) is a span.

Let $\mathbf{(}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$spans K over Fand $\mathit{w}\mathbf{\in }\mathit{K}$be any element.

Since $\mathbf{(}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$spans K.

Therefore $\mathit{w}\mathbf{\in }\mathit{K}$can be written as linear combination of$\mathbf{(}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$. Thus

$\mathit{w}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{c}}_{\mathbf{n}}{\mathit{x}}_{\mathbf{n}}$. Where ${\mathit{c}}_{\mathbf{i}}\mathbf{\in }\mathit{F}$for any $\mathit{v}\mathbf{\in }\mathit{K}$.

$\mathit{v}\mathbf{=}{\mathit{\alpha }}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}\mathbf{+}{\mathit{\alpha }}_{\mathbf{2}}{\mathit{v}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Add and subtract $\mathit{w}\mathbf{\in }\mathit{F}$in the equation (1).

$$\begin{gathered} \mathit{v}\mathbf{=}{\mathit{\alpha }}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}\mathbf{+}{\mathit{\alpha }}_{\mathbf{2}}{\mathit{v}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{\alpha }}_{\mathbf{n}}{\mathit{v}}_{\mathbf{n}}\mathbf{-}\mathit{w}\mathbf{+}\mathit{w} \\ \mathbf{}\mathbf{}\mathbf{=}{\mathit{\alpha }}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}\mathbf{+}{\mathit{\alpha }}_{\mathbf{2}}{\mathit{v}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{\alpha }}_{\mathbf{n}}{\mathit{v}}_{\mathbf{n}}\mathbf{+}\mathit{w}\mathbf{-}\left(c_1{v}_1+c_2{v}_2+......c_n{v}_n\right) \\ \mathbf{}\mathbf{}\mathbf{=}\left({\alpha }_1-c_1\right){\mathit{v}}_{\mathbf{1}}\mathbf{+}\left({\alpha }_2-c_2\right){\mathit{v}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left({\alpha }_n-c_n\right){\mathit{v}}_{\mathbf{n}}\mathbf{+}\mathit{w} \end{gathered}$$

Put ${\mathit{\beta }}_{\mathbf{i}}\mathbf{=}{\mathit{\alpha }}_{\mathbf{i}}\mathbf{-}{\mathit{c}}_{\mathbf{i}}$

$\mathit{v}\mathbf{=}{\mathit{\beta }}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{v}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{\beta }}_{\mathbf{n}}{\mathit{v}}_{\mathbf{n}}\mathbf{+}\mathit{w}$

Thus $\mathit{v}\mathbf{\in }\mathit{K}$is written as a linear combination of elements of set$\mathbf{(}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$.

Therefore role="math" localid="1656921162831" $\mathbf{(}\mathit{w}\mathbf{,}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{)}$also spans K.