Question

If${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$are any three distinct vectors in ${\mathrm{ℝ}}^3$, then there must be a linear transformation T from ${\mathrm{ℝ}}^{3}to{\mathrm{ℝ}}^3$such that$T\left({\overrightarrow{v}}_1\right)={\overrightarrow{e}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{e}}_{2}andT\left({\overrightarrow{v}}_3\right)={\overrightarrow{e}}_3$.

Short answer

The above statement is false.

If ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$are any three distinct vectors in ${\mathrm{ℝ}}^3$, then there is no linear transformation T from ${\mathrm{ℝ}}^3\text{to}{\mathrm{ℝ}}^3$such that $T\left({\overrightarrow{v}}_1\right)={\overrightarrow{e}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{e}}_2\text{and}T\left({\overrightarrow{v}}_3\right)={\overrightarrow{e}}_3$.

Step-by-step solution

Mentioning concept

Let T be a linear transformation then the linear transformation T maps the basis vector onto the basis vector.

Considering three distinct vectors v→1,v→2,v→3

Let us consider three distinct vectors${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$ such that

${\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ 2\\ 7\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}-2\\ 1\\ -4\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$

Now form the matrix whose columns are given vectors, and reduce to echelon form using elementary row operations.

$\left[\begin{array}{ccc}1& -2& 1\\ 2& 1& -1\\ 7& -4& 1\end{array}\right]\approx \left[\begin{array}{ccc}1& -2& 1\\ 0& 5& -3\\ 0& 10& -6\end{array}\right]\approx \left[\begin{array}{ccc}1& -2& 1\\ 0& 5& -3\\ 0& 0& 0\end{array}\right]$

Since the echelon matrix has a row zero, the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$are linearly dependent.

Hence three distinct vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$don’t form a basis.

So, there will be no linear transformation T from ${\mathrm{ℝ}}^3\text{to}{\mathrm{ℝ}}^3$such that

$T\left({\overrightarrow{v}}_1\right)={\overrightarrow{e}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{e}}_2\text{and}T\left({\overrightarrow{v}}_3\right)={\overrightarrow{e}}_3.$

Final Answer

Since it is given that ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$are any three distinct vectors in${\mathrm{ℝ}}^3$ , but the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$could be linearly dependent vectors and don’t form a basis; hence there is no linear transformation T from ${\mathrm{ℝ}}^3\text{to}{\mathrm{ℝ}}^3$for the vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$such that$T\left({\overrightarrow{v}}_1\right)={\overrightarrow{e}}_1,T\left({\overrightarrow{v}}_2\right)={\overrightarrow{e}}_2\text{and}T\left({\overrightarrow{v}}_3\right)={\overrightarrow{e}}_3.$