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.

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
asked 2020-11-07
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.

Answers (1)

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.
0

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