# Prove that displaystyle{T}{left({x},{y}right)}={left({3}{x}+{y},{2}{y},{x}-{y}right)} defines a linear transformation displaystyle{T}:mathbb{R}^{2}tomathbb{R}^{3}. Give the full and correct answer.

Question
Vectors and spaces
Prove that $$\displaystyle{T}{\left({x},{y}\right)}={\left({3}{x}+{y},{2}{y},{x}-{y}\right)}$$ defines a linear transformation $$\displaystyle{T}:\mathbb{R}^{2}\to\mathbb{R}^{3}$$. Give the full and correct answer.

2020-11-08
Let $$\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)},{\left({x}_{{2}},{x}_{{2}}\right)}\in\mathbb{R}^{2}{\quad\text{and}\quad}\le{t}{c}_{{1}},{c}_{{2}}\in\mathbb{R}$$. To prove that T is a linear transformation, we must prove that
$$\displaystyle{T}{\left({c}_{{1}}{\left({x}_{{1}},{y}_{{1}}\right)}+{c}_{{2}}{\left({x}_{{2}},{y}_{{2}}\right)}\right)}={c}_{{1}}{T}{\left({x}_{{1}},{y}_{{1}}\right)}+{c}_{{2}}{T}{\left({x}_{{2}},{y}_{{2}}\right)}$$
This is done by a direct computation:
$$\displaystyle{T}{\left({c}_{{1}}{\left({x}_{{1}},{y}_{{1}}\right)}+{c}_{{2}}{\left({x}_{{2}},{y}_{{2}}\right)}\right)}$$
$$\displaystyle={T}{\left({c}_{{1}}{x}_{{1}},{c}_{{2}}{x}_{{2}}\right)}+{\left({c}_{{2}}{x}_{{2}},{c}_{{2}}{x}_{{2}}\right)}$$
$$\displaystyle={T}{\left({c}_{{1}}{x}_{{1}}{c}_{{2}}{x}_{{2}},{c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\right)}$$
$$\displaystyle={\left({3}{\left({c}_{{1}}{x}_{{1}}+{c}_{{2}}{x}_{{2}}\right)}+{\left({c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\right)},{2}{\left({c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\right)},{\left({c}_{{1}}{x}_{{1}}+{c}_{{2}}{x}_{{2}}\right)}-{\left({c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\right)}\right)}$$
$$\displaystyle={\left({3}{c}_{{1}}{x}_{{1}}+{3}{c}_{{2}}{x}_{{2}}+{c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}},{2}{c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}},{c}_{{1}}{x}_{{1}}+{c}_{{2}}{x}_{{2}}-{c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\right)}$$
$$\displaystyle={\left({3}{c}_{{1}}{x}_{{1}}+{c}_{{1}}{y}_{{1}},{2}{c}_{{1}}{y}_{{1}},{c}_{{1}}{x}_{{1}}-{c}_{{1}}{y}_{{1}}\right)}+{\left({3}{c}_{{2}}{x}_{{2}}+{c}_{{2}}{y}_{{2}},{2}{c}_{{2}}{y}_{{2}},{c}_{{2}}{x}_{{2}}-{c}_{{2}}{y}_{{2}}\right)}$$
$$\displaystyle={c}_{{1}}{\left({3}{x}_{{1}}+{y}_{{1}},{2}{y}_{{1}},{x}_{{1}}-{y}_{{1}}\right)}+{c}_{{2}}{\left({3}{x}_{{2}}+{y}_{{2}},{2}{y}_{{2}},{x}_{{2}}-{y}_{{2}}\right)}$$
$$\displaystyle={c}_{{1}}{T}{\left({x}_{{1}},{y}_{{1}}\right)}+{c}_{{2}}{T}{\left({x}_{{2}},{y}_{{2}}\right)}$$
Therefore, T is linear.

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$$\displaystyle{1}.{f{{\left({x}_{{1}},{x}_{{2}}\right)}}}={\left({2}{x}_{{1}}-{x}_{{2}},{3}{x}_{{1}}+{x}_{{2}}\right)}$$
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$$\displaystyle{4}.{f{{\left({x},{y}\right)}}}={\left({2}{x}+{y},-{3}{x}+{5}{y}\right)}$$
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$$\displaystyle{6}.{L}{\left({x},{y}\right)}={\left({x},{x}+{y},-{y}\right)}$$