Whether the function is a linear transformation or not. T : R^{2} rightarrow R^{3}, T(x,y)=(sqrt{x},xy,sqrt{y})

Whether the function is a linear transformation or not. T : R^{2} rightarrow R^{3}, T(x,y)=(sqrt{x},xy,sqrt{y})

Question
Transformation properties
asked 2021-03-06
Whether the function is a linear transformation or not.
\(\displaystyle{T}\ :\ {R}^{{{2}}}\rightarrow{R}^{{{3}}},{T}{\left({x},{y}\right)}={\left(\sqrt{{{x}}},{x}{y},\sqrt{{{y}}}\right)}\)

Answers (1)

2021-03-07

Calculation:
The function is defined as,
\(\displaystyle{T}{\left({x},{y}\right)}={\left(\sqrt{{{x}}},{x}{y},\sqrt{{{y}}}\right)}\)
Assume two general vectors \(\displaystyle{u}={\left({u}_{{{1}}},{u}_{{{2}}}\right)}\) and \(\displaystyle{v}={\left({v}_{{{1}}},{v}_{{{2}}}\right)}\)
Then \(\displaystyle{u}+{v}={\left({u}_{{{1}}}+{v}_{{{1}}},{u}_{{{2}}}+{v}_{{{2}}}\right)}\)
\(\displaystyle{c}{u}={\left({c}{u}_{{{1}}},{c}{u}_{{{2}}}\right)}\)
The function is a linear transformation if it satisfies the two properties as mentioned in the approach part.
Compute \(\displaystyle{T}{\left({u}+{v}\right)}\) and \(\displaystyle{T}{\left({u}\right)}+{T}{\left({v}\right)}\) as,
\(\displaystyle{T}{\left({u}+{v}\right)}={T}{\left({u}_{{{1}}}+{v}_{{{1}}},{u}_{{{2}}}+{v}_{{{2}}}\right)}\)
\(\displaystyle={\left(\sqrt{{{u}_{{{1}}}+{v}_{{{1}}}}},{\left({u}_{{{1}}}+{v}_{{{1}}}\right)}{\left({u}_{{{2}}}+{v}_{{{2}}}\right)},\sqrt{{{u}_{{{2}}}+{v}_{{{2}}}}}\right)}\)
\(\displaystyle={\left(\sqrt{{{u}_{{{1}}}+{v}_{{{1}}}}},{\left({u}_{{{1}}}{u}_{{{2}}}+{u}_{{{1}}}{y}_{{{2}}}+{u}_{{{2}}}{v}_{{{1}}}+{v}_{{{1}}}{v}_{{{2}}}\right)},{\left(\sqrt{{{u}_{{{2}}}+{v}_{{{2}}}}}\right)}\right.}\)
\(\displaystyle{T}{\left({u}\right)}+{T}{\left({v}\right)}={T}{\left({u}_{{{1}}},{u}_{{{2}}}\right)}+{T}{\left({v}_{{{1}}},{v}_{{{2}}}\right)}\)
PSK\(=(\sqrt{u_{1}},u_{1}u_{2},\sqrt{u_{2}})+(\sqrt{v_{1}},v_{1}v_{2},\sqrt{v_{2}})\)
\(\displaystyle={\left(\sqrt{{{u}_{{{1}}}}}+\sqrt{{{v}_{{{1}}}}},{u}_{{{1}}}{u}_{{{2}}}+{v}_{{{1}}}{v}_{{{2}}},\sqrt{{{u}_{{{2}}}}}+\sqrt{{{v}_{{{2}}}}}\right.}\)
Since \(\displaystyle{T}{\left({u}+{v}\right)}\ne{q}{T}{\left({u}\right)}+{T}{\left({v}\right)}\), the first property is not satisfied.
Therefore, the function is not a linear transformation.

0

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