Question

If a subspace V of$R^3$ contains the standard vectors ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3$then V must be $R^3$.

Short answer

The above statement is true.

If a subspace V of $R^3$contains the standard vectors ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3$then V must be$R^3.$

Step-by-step solution

Condition of a subspace to be equal to the vector space

Let V be a subspace of a vector space$R^n\left(F\right)$ , then the subspace V of $R^n\left(F\right)$is equal to the vector space$R^n\left(F\right)$ if dim(V) = dim($R^n\left(F\right)$) = n.

To show dim(V) = dim(R3)

Since the subspace V of $R^3$contains the standard vectors${\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3$ where

${\overrightarrow{e}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\text{,}{\overrightarrow{e}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\text{,}{\overrightarrow{e}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

And the standard vectors ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3$are linearly independent, so, these vectors form the bases of subspace V.

Thus, dimension of V = 3 = dimension of$R^3.$

Hence, V =$R^3.$

Final Answer

Since, dimension of V = 3 = dimension of$R^3.$

$\Rightarrow V=R^3.$

Hence, if a subspace V of $R^3$contains the standard vectors ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3$then V must be$R^3.$