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

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

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.

### Relevant Questions

Whether the function is a linear transformation or not.
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Whether T is a linear transformation, that is $$\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}$$ defined by $$\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}$$
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To show:
The set $$\displaystyle{\left\lbrace{T}{\left({x}_{{1}}\right)},\ \ldots\ ,{T}{\left({x}_{{k}}\right)}\right\rbrace}$$ is a linearly independent subset of $$\displaystyle{R}^{{{m}}}$$
Given:
Let $$\displaystyle{T}\ :\ {T}\ :\ {R}^{{{n}}}\rightarrow{R}^{{{m}}}$$ be a linear transformation with nulity zero. If $$\displaystyle{S}={\left\lbrace{x}_{{{1}}},\ \cdots\ \ ,{x}_{{{k}}}\right\rbrace}$$ is a linearly independent subset of $$\displaystyle{R}^{{{n}}}.$$
To determine:
To prove:
The function $$\displaystyle{T}:{V}\rightarrow{W}$$ is linear transformation if and only if
$$\displaystyle{T}{\left({a}{u}\ +\ {b}{v}\right)}={a}{T}{\left({u}\right)}\ +\ {b}{T}{\left({v}\right)}$$ for all vectors u and v and all scalars a and b.
Find the linear or affine transformations that satisty the desired properties and write it in the form $$T(x) = Ax\ +\ b:$$
The transformation $$T\ :\ \mathbb{R}^{2}\ \rightarrow\ \mathbb{R}^{2}$$ sending the origin to itself and a triangle of vertices
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$$(\sqrt{\frac{2}{2}},\ \sqrt{\frac{2}{2}}),\ (- \sqrt{\frac{2}{2}},\ \sqrt{2}).$$
Let $$\displaystyle\le{f}{t}{\left\lbrace{v}_{{{1}}},\ {v}_{{{2}}},\dot{{s}},\ {v}_{{{n}}}{r}{i}{g}{h}{t}\right\rbrace}$$ be a basis for a vector space V. Prove that if a linear transformation $$\displaystyle{T}\ :\ {V}\rightarrow\ {V}$$ satisfies $$\displaystyle{T}{\left({v}_{{{1}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n},$$ then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that $$\displaystyle{T}{\left({v}\right)}={0}$$ for every vector v in V.
(i) Let v be an arbitrary vector in V such that $$\displaystyle{v}={c}_{{{1}}}\ {v}_{{{1}}}\ +\ {c}_{{{2}}}\ {v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {v}_{{{n}}}.$$
(ii) Use the definition and properties of linear transformations to rewrite $$\displaystyle{T}\ {\left({v}\right)}$$ as a linear combination of $$\displaystyle{T}\ {\left({v}_{{{1}}}\right)}$$.
(iii) Use the fact that $$\displaystyle{T}\ {\left({v}_{{i}}\right)}={0}$$ to conclude that $$\displaystyle{T}\ {\left({v}\right)}={0}$$, making T the zero tranformation.
Prove whether $${f}:\mathbb{R}\to\mathbb{R}\ \text{defined by}\ f{{\left({x}\right)}}={4}{x}-{2}$$ is a linear transformation.
Prove whether $${f}:\mathbb{R}\to\mathbb{R}\ \text{defined by}\ f{{\left({x}\right)}}={2}{x}$$ is a linear transformation.
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$$\displaystyle\text{Let A be an}\ {n}\ \times\ {n}\ \text{matrix and suppose that}\ {L}:{M}_{{\cap}}\ \rightarrow\ {M}_{{\cap}}\ \text{is defined by}{L}{\left({x}\right)}={A}{X},\text{for}\ {X}\in{M}_{{\cap}}.\text{Show that L is a linear transformation.}$$