# Use a counterexample to show that the statement is false. {T}:{R}^{2}to{R}^{2},{T}{left({x}_{{2}},{x}_{{2}}right)}={left({x}_{{1}}+{4},{x}_{{2}}right)} is a linear transformation?

Use a counterexample to show that the statement is false.
$T:{R}^{2}\to {R}^{2},T\left({x}_{2},{x}_{2}\right)=\left({x}_{1}+4,{x}_{2}\right)$ is a linear transformation?
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Approach:
Let V and W be vector spaces. The function $T:V\to W$ is a linear transformation of V into W when the two properties below are true for all u and v in V and for any scalar c.
$T\left(u+v\right)=T\left(u\right)+T\left(v\right)$
$T\left(cu\right)=cT\left(u\right)$
Calculation:
Assume the two vectors $u=\left({u}_{1},{u}_{2}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}v=\left({v}_{1},{v}_{2}\right).$
The sum of the two vectors is,
$u+v=\left({u}_{1},{u}_{2}\right)+\left({v}_{1},{v}_{2}\right)$
$=\left({u}_{1}+{v}_{1},{u}_{2}+{v}_{2}\right)$
Apply transformation on both side of the above equation.
$T\left(u+v\right)=T\left({u}_{1}+{v}_{1},{u}_{2}+{v}_{2}\right)$
$=\left({u}_{1}+{v}_{1}+4,{u}_{2}+{v}_{2}\right)\dots \left(1\right)$
The sum of the two transformations is,
$T\left(u\right)+T\left(v\right)=T\left({u}_{1},{u}_{2}\right)+T\left({v}_{1},{v}_{2}\right)$
$=\left({u}_{1}+4,{u}_{2}\right)+\left({v}_{1}+4,{v}_{2}\right)$
$=\left({u}_{1}+{v}_{1}+8,{u}_{2}+{v}_{2}\right)\dots \left(2\right)$
From equation (1) and equation (2),
$T\left(u+v\right)\ne T\left(u\right)+T\left(v\right)$
From the above result, the function $T\left({x}_{1},{x}_{2}\right)=\left({x}_{1}+4,{x}_{2}\right)$ is not a linear transformation.
Therefore, the statement is false.