banganX

2020-11-07

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

StrycharzT

Let $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{x}_{2}\right)\in {\mathbb{R}}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\le t{c}_{1},{c}_{2}\in \mathbb{R}$. To prove that T is a linear transformation, we must prove that
$T\left({c}_{1}\left({x}_{1},{y}_{1}\right)+{c}_{2}\left({x}_{2},{y}_{2}\right)\right)={c}_{1}T\left({x}_{1},{y}_{1}\right)+{c}_{2}T\left({x}_{2},{y}_{2}\right)$
This is done by a direct computation:
$T\left({c}_{1}\left({x}_{1},{y}_{1}\right)+{c}_{2}\left({x}_{2},{y}_{2}\right)\right)$
$=T\left({c}_{1}{x}_{1},{c}_{2}{x}_{2}\right)+\left({c}_{2}{x}_{2},{c}_{2}{x}_{2}\right)$
$=T\left({c}_{1}{x}_{1}{c}_{2}{x}_{2},{c}_{1}{y}_{1}+{c}_{2}{y}_{2}\right)$
$=\left(3\left({c}_{1}{x}_{1}+{c}_{2}{x}_{2}\right)+\left({c}_{1}{y}_{1}+{c}_{2}{y}_{2}\right),2\left({c}_{1}{y}_{1}+{c}_{2}{y}_{2}\right),\left({c}_{1}{x}_{1}+{c}_{2}{x}_{2}\right)-\left({c}_{1}{y}_{1}+{c}_{2}{y}_{2}\right)\right)$
$=\left(3{c}_{1}{x}_{1}+3{c}_{2}{x}_{2}+{c}_{1}{y}_{1}+{c}_{2}{y}_{2},2{c}_{1}{y}_{1}+{c}_{2}{y}_{2},{c}_{1}{x}_{1}+{c}_{2}{x}_{2}-{c}_{1}{y}_{1}+{c}_{2}{y}_{2}\right)$

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