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

Question
Transformation properties
asked 2021-01-08
Whether the function is a linear transformation or not.
\(\displaystyle{T}\ :\ {R}^{{{2}}}\rightarrow{R}^{{{2}}},{T}{\left({x},{y}\right)}={\left({x},{y}^{{{2}}}\right)}\)

Answers (1)

2021-01-09
Calculation:
The function is defined as,
\(\displaystyle{T}{\left({x},{y}\right)}={\left({x},{y}^{{{2}}}\right)}\)
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
\(\displaystyle{T}{\left({u}+{v}\right)}={T}{\left({u}_{{{1}}}+{v}_{{{1}}},\ {u}_{{{2}}}+{v}_{{{2}}}\right)}\)
\(\displaystyle={\left({u}_{{{1}}}+{v}_{{{1}}},\ {\left({u}_{{{2}}}+{v}_{{{2}}}\right)}^{{{2}}}\right)}\)
\(\displaystyle={\left({u}_{{{1}}}+{v}_{{{1}}},\ {{u}_{{{2}}}^{{{2}}}}+{{v}_{{{2}}}^{{{2}}}}\ +\ {2}{u}_{{{2}}}{v}_{{{2}}}\right)}\)
\(\displaystyle{T}{\left({u}\right)}+{T}{\left({v}\right)}={T}{\left({u}_{{{1}}},\ {u}_{{{2}}}\right)}+{T}{\left({v}_{{{1}}}\ ,{v}_{{{2}}}\right)}\)
\(\displaystyle={\left({u}_{{{1}}},\ {{u}_{{{2}}}^{{{2}}}}\right)}+{\left({v}_{{{1}}},\ {{v}_{{{2}}}^{{{2}}}}\right)}\)
\(\displaystyle={\left({u}_{{{1}}}+{v}_{{{1}}},\ {{u}_{{{2}}}^{{{2}}}}+{{v}_{{{2}}}^{{{2}}}}\right)}\)
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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