Question

Consider the vectors${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}$in${\mathbf{ℝ}}^{\mathbf{n}}$. Is span$\left({\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}\right)$ necessarily a subspace of${\mathbf{ℝ}}^{\mathbf{n}}$ ? Justify your answer.

Short answer

The span $\left({\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,.......,{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}\right)$is a subspace of ${\mathrm{ℝ}}^n$.

Step-by-step solution

Consider the set.

A subset W of the vector space ${\mathrm{ℝ}}^n$is called a (linear) subspace of ${\mathrm{ℝ}}^n$if it has the following three properties:

a. W contains the zero vector in ${\mathrm{ℝ}}^n$.

b. W is closed under addition: If ${\overrightarrow{w}}_1$and ${\overrightarrow{w}}_2$are both in W, then so is ${\overrightarrow{w}}_1+{\overrightarrow{w}}_2$.

c. W is closed under scalar multiplication: If $\overrightarrow{w}$is in W and k is an arbitrary scalar, then $k\overrightarrow{w}$is in W.

The span $\left({\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,.......,{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}\right)$should satisfy all the above conditions.

Check for first condition.

The first condition is,

$\left({\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,.......,{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}\right)$can be written as,

$\therefore \overrightarrow{0}\in W$

The first condition holds good.

Check for second condition.

The second condition is,

$\overrightarrow{0}=0\left({\overrightarrow{v}}_1+{\overrightarrow{v}}_2\right)$

${\overrightarrow{v}}_1+{\overrightarrow{v}}_2\in W$

The second condition holds good.

Check for third condition

The third condition is,

The linear combination of span $\left({\overrightarrow{v}}_1,{\overrightarrow{v}}_2,.........,{\overrightarrow{v}}_m\right)$is still a part of the span of span

The third condition also holds good.

Final answer.

The span $\left({\overrightarrow{v}}_1,{\overrightarrow{v}}_2,.........,{\overrightarrow{v}}_m\right)$is a subspace of ${\mathrm{ℝ}}^n$.