Question

Letbe the basis of${\mathrm{ℝ}}^n$consisting of the vectors$\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{2}}}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{3}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{n}}}\mathbf{,}$and letlocalid="1660636061360" $\mathfrak{R}$be some other basis of ${\mathrm{ℝ}}^n$. Islocalid="1660645467599" ${\left[\overrightarrow{v_1}\right]}_{\mathbf{R}}\mathbf{,}{\left[\overrightarrow{v_2}\right]}_{\mathbf{R}}\mathbf{,}{\left[\overrightarrow{v_3}\right]}_{\mathbf{R}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\left[\overrightarrow{v_n}\right]}_{\mathbf{R}}$a basis of${\mathbf{ℝ}}^{\mathbf{n}}$ as well?

Explain.

Short answer

If $؏$be the basis of ${\mathrm{ℝ}}^n$consisting of the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,...,{\overrightarrow{v}}_n$, and let be some other basis of ${\mathrm{ℝ}}^n$then the vectors ${\left[\overrightarrow{v_1}\right]}_{\mathbf{R}}\mathbf{,}{\left[\overrightarrow{v_2}\right]}_{\mathbf{R}}\mathbf{,}{\left[\overrightarrow{v_3}\right]}_{\mathbf{B}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\left[\overrightarrow{v_n}\right]}_{\mathbf{B}}$forms a basis of${\mathrm{ℝ}}^n$ as well.

Step-by-step solution

Mentioning concept

Given $؏=\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...,{\overrightarrow{v}}_n\right\}$is the basis of ${\mathrm{ℝ}}^n$. Then for any scalars $a_1,a_2,\dots ,a_n$we have

$$\begin{gathered} a_1\overrightarrow{v_1}+a_2{\overrightarrow{v}}_2+\dots +a_n\overrightarrow{v_n}=0 \\ \Rightarrow a_1=a_2=...=a_n=0 \end{gathered}$$

Let$\mathfrak{R}=\left\{{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,...,{\overrightarrow{w}}_n\right\}$ be another of${\mathrm{ℝ}}^n$. Then for any scalars$b_1,b_2,...,b_n$ we have

$$\begin{gathered} b_1\overrightarrow{w_1}+b_2{\overrightarrow{w}}_2+\dots +b_n\overrightarrow{w_n}=0 \\ \Rightarrow b_1=b_2=...=b_n=0 \end{gathered}$$

 Linearity of coordinates

If $؏$be a basis of a subspace V of ${\mathrm{ℝ}}^n$then

1. ${\left[\overrightarrow{x}+\overrightarrow{y}\right]}_{؏}={\left[\overrightarrow{x}\right]}_{؏}+{\left[\overrightarrow{y}\right]}_{؏}$for all vectors role="math" localid="1660646739789" $\overrightarrow{\mathrm{x}}\mathrm{and}\overrightarrow{\mathrm{y}}$in V (1)
2. ${\left[k\overrightarrow{x}\right]}_{؏}=k{\left[\overrightarrow{x}\right]}_{؏}$for all vectors$\overrightarrow{x}$ in V and all scalars k. (2)

To prove [v1→]R,[v2→]R,[v3→]R,....,[vn→]R is a basis of ℝn

Let us suppose the vectors ${\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}},...,{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}$are linearly dependent. Then there exists scalars$c_1,c_2,...,c_n$, not all of them 0, such that

$c_1{\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}}+c_2{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}}+c_3{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}}+...+c_n{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}=0$

Therefore from equation (2), we have

$c_1{\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}}+c_2{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}}+c_3{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}}+...+c_n{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}=0$

Now from equation (1), we get

${\left[c_1{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}}+{\left[c_1{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}}+{\left[c_3{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}}+...+{\left[c_1{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}=0$

We know that, if${\left[x\right]}_{\mathfrak{R}}=0\Rightarrow \overrightarrow{x}=0$.

Therefore

$c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+c_3{\overrightarrow{v}}_3+...+c_n{\overrightarrow{v}}_n=0$

Since $؏=\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...,\overrightarrow{v_n}\right\}$is a basis of ${\mathrm{ℝ}}^n$, then the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,...,{\overrightarrow{v}}_n$are linearly independent.

$\Rightarrow c_1=c_2=...=c_n=0$

Since we have supposed that the vectors ${\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}},...,{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}$are linearly dependent such that

$c_1{\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}}+c_2{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}}+c_3{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}}+...+c_n{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}=0$

But we have

$c_1=c_2=...=c_n=0$

Therefore the vectors${\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}},...,{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}$are linearly independent and thus form a basis of${\mathrm{ℝ}}^n$

 Final Answer

If $؏$be the basis of ${\mathrm{ℝ}}^n$consisting of the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,...,{\overrightarrow{v}}_n$, and let $\mathfrak{R}$be some other basis of ${\mathrm{ℝ}}^n$then the vectors${\left[{\overrightarrow{\mathrm{v}}}_1\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_2\right]}_{\mathfrak{R}},{\left[{\overrightarrow{\mathrm{v}}}_3\right]}_{\mathfrak{R}},...,{\left[{\overrightarrow{\mathrm{v}}}_{\mathrm{n}}\right]}_{\mathfrak{R}}$forms a basis of${\mathrm{ℝ}}^n$ as well.