 # Whether the function is a linear transformation or not. T : R^{2} rightarrow R^{3}, T(x,y)=(sqrt{x},xy,sqrt{y}) alesterp 2021-03-06 Answered
Whether the function is a linear transformation or not.
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Calculation:
The function is defined as,
$T\left(x,y\right)=\left(\sqrt{x},xy,\sqrt{y}\right)$
Assume two general vectors $u=\left({u}_{1},{u}_{2}\right)$ and $v=\left({v}_{1},{v}_{2}\right)$
Then $u+v=\left({u}_{1}+{v}_{1},{u}_{2}+{v}_{2}\right)$
$cu=\left(c{u}_{1},c{u}_{2}\right)$
The function is a linear transformation if it satisfies the two properties as mentioned in the approach part.
Compute $T\left(u+v\right)$ and $T\left(u\right)+T\left(v\right)$ as,
$T\left(u+v\right)=T\left({u}_{1}+{v}_{1},{u}_{2}+{v}_{2}\right)$
$=\left(\sqrt{{u}_{1}+{v}_{1}},\left({u}_{1}+{v}_{1}\right)\left({u}_{2}+{v}_{2}\right),\sqrt{{u}_{2}+{v}_{2}}\right)$
$=\left(\sqrt{{u}_{1}+{v}_{1}},\left({u}_{1}{u}_{2}+{u}_{1}{y}_{2}+{u}_{2}{v}_{1}+{v}_{1}{v}_{2}\right),\left(\sqrt{{u}_{2}+{v}_{2}}\right)$
$T\left(u\right)+T\left(v\right)=T\left({u}_{1},{u}_{2}\right)+T\left({v}_{1},{v}_{2}\right)$
$=\left(\sqrt{{u}_{1}},{u}_{1}{u}_{2},\sqrt{{u}_{2}}\right)+\left(\sqrt{{v}_{1}},{v}_{1}{v}_{2},\sqrt{{v}_{2}}\right)$
$=\left(\sqrt{{u}_{1}}+\sqrt{{v}_{1}},{u}_{1}{u}_{2}+{v}_{1}{v}_{2},\sqrt{{u}_{2}}+\sqrt{{v}_{2}}$
Since $T\left(u+v\right)\ne qT\left(u\right)+T\left(v\right)$, the first property is not satisfied.
Therefore, the function is not a linear transformation.