Get the solution to "Whether the function is a linear transformation or..."

High School
Geometry
High school geometry
Transformation properties

alesterp

Answered question

2021-03-06

Whether the function is a linear transformation or not.

T : R^{2} \to R^{3}, T (x, y) = (\sqrt{x}, x y, \sqrt{y})

Answer & Explanation

Aubree Mcintyre

Skilled2021-03-07Added 73 answers

Calculation:
The function is defined as,
$T (x, y) = (\sqrt{x}, x y, \sqrt{y})$
Assume two general vectors $u = (u_{1}, u_{2})$ and $v = (v_{1}, v_{2})$
Then $u + v = (u_{1} + v_{1}, u_{2} + v_{2})$
$c u = (c u_{1}, c u_{2})$
The function is a linear transformation if it satisfies the two properties as mentioned in the approach part.
Compute $T (u + v)$ and $T (u) + T (v)$ as,
$T (u + v) = T (u_{1} + v_{1}, u_{2} + v_{2})$
$= (\sqrt{u_{1} + v_{1}}, (u_{1} + v_{1}) (u_{2} + v_{2}), \sqrt{u_{2} + v_{2}})$
$= (\sqrt{u_{1} + v_{1}}, (u_{1} u_{2} + u_{1} y_{2} + u_{2} v_{1} + v_{1} v_{2}), (\sqrt{u_{2} + v_{2}})$
$T (u) + T (v) = T (u_{1}, u_{2}) + T (v_{1}, v_{2})$
$= (\sqrt{u_{1}}, u_{1} u_{2}, \sqrt{u_{2}}) + (\sqrt{v_{1}}, v_{1} v_{2}, \sqrt{v_{2}})$
$= (\sqrt{u_{1}} + \sqrt{v_{1}}, u_{1} u_{2} + v_{1} v_{2}, \sqrt{u_{2}} + \sqrt{v_{2}}$
Since $T (u + v) \neq q T (u) + T (v)$ , the first property is not satisfied.
Therefore, the function is not a linear transformation.

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