Question

If${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,...,{\overrightarrow{v}}_{n}and{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,{\overrightarrow{w}}_3,...,{\overrightarrow{w}}_n$ are two bases of ${\mathrm{ℝ}}^n$, then there exists a linear transformation T from${\mathrm{ℝ}}^n\text{to}{\mathrm{ℝ}}^n$ such that $T\left({\overrightarrow{v}}_1\right)={\overrightarrow{w}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{w}}_2,...,T\left({\overrightarrow{v}}_n\right)={\overrightarrow{w}}_n$.

Short answer

The above statement is true.

If${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,...,{\overrightarrow{v}}_n\text{and}{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,{\overrightarrow{w}}_3,...,{\overrightarrow{w}}_n$ are two bases of${\mathrm{ℝ}}^n$ , then there exists a linear transformation T from ${\mathrm{ℝ}}^n\text{to}{\mathrm{ℝ}}^n$such that $T\left({\overrightarrow{v}}_1\right)={\overrightarrow{w}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{w}}_2,...,T\left({\overrightarrow{v}}_n\right)={\overrightarrow{w}}_n$.

Step-by-step solution

Considering a mapping

Let $\overrightarrow{x}\in {\mathrm{ℝ}}^n.$ Then

$$\begin{gathered} \overrightarrow{x}=\sum _{i=1}^n{a}_i{v}_i \\ \text{where}{a}_1,a_2,...,a_n\text{are unique scalars} \end{gathered}$$

Define a map

$$\begin{gathered} T:{\mathrm{ℝ}}^n\to {\mathrm{ℝ}}^n\text{by} \\ \text{T(}x)=\sum _{i=1}^n{a}_i{w}_i\text{.} \end{gathered}$$

To show T is linear

Suppose that$u,v\in {\mathrm{ℝ}}^n\text{and}d\in F.$

Then we may write

$\overrightarrow{u}=\sum _{i=1}^n{b}_i{v}_i\text{and}\overrightarrow{v}=\sum _{i=1}^n{c}_i{v}_i$

for scalars$b_1,b_2,....,b_n\text{and}{c}_1,c_2...,c_n$ .

Thus,

$du+v=\sum _{i=n}^n(d{b}_i+c_i)v_i$

$\Rightarrow T(du+v)=\sum _{i=1}^n(d{b}_i+c_i)w_i.$

$\Rightarrow T(du+v)=d\sum _{i=1}^n(b_i{w}_i)+\sum _{i=n}^n{c}_i{w}_i=dT(u)+T(v).$

Clearly,$T\left(v_i\right)=w_i\text{for}i=1,2,...,n.$

T is unique

Suppose $U:{\mathrm{ℝ}}^n\to {\mathrm{ℝ}}^n$is linear and$U\left(v_I\right)=w_I\text{for}i=1,2,...,n.$

Then for $\overrightarrow{x}\in {\mathrm{ℝ}}^n$with

$$\begin{gathered} \overrightarrow{x}=\sum _{i=1}^n{a}_i{v}_i \\ \text{where}{a}_1,a_2,...,a_n\text{are unique scalars.} \end{gathered}$$

We have

$U\left(x\right)=\sum _{i=1}^n{a}_{i}U\left(v_i\right)=\sum _{i=1}^n{a}_i{w}_i=T\left(x\right)\text{.}$

Hence, U =T.

Final Answer

If ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,...,{\overrightarrow{v}}_n\text{and}{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,{\overrightarrow{w}}_3,...,{\overrightarrow{w}}_n$are two bases of ${\mathrm{ℝ}}^n$ , then there exists a unique linear transformation T from${\mathrm{ℝ}}^n\text{to}{\mathrm{ℝ}}^n$ such that $T\left({\overrightarrow{v}}_1\right)={\overrightarrow{w}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{w}}_2,...,T\left({\overrightarrow{v}}_n\right)={\overrightarrow{w}}_n$.