Ikunupe6v

2021-12-20

How to determine which of the following transformations are linear transformations?
Determine which of the following transformations are linear transformations
A. The transformation ${T}_{1}$ defined by ${T}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)=\left({x}_{1},0,{x}_{3}\right)$
B. The transformation ${T}_{2}$ defined by ${T}_{2}\left({x}_{1},{x}_{2}\right)=\left(2{x}_{1}-3{x}_{2},{x}_{1}+4,5{x}_{2}\right)$
C. The transformation ${T}_{3}$ defined by ${T}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right)=\left({x}_{1},{x}_{2},-{x}_{3}\right)$
D. The transformation ${T}_{4}$ defined by ${T}_{4}\left({x}_{1},{x}_{2},{x}_{3}\right)=\left(1,{x}_{2},{x}_{3}\right)$
E. The transformation ${T}_{5}$ defined by ${T}_{5}\left({x}_{1},{x}_{2}\right)=\left(4{x}_{1}-2{x}_{2},3|{x}_{2}|\right)$

Anzante2m

Expert

To test whether T is a linear transformation, you must verify that. for some vectors a and b and some constant c
$T\left(a+b\right)=T\left(a\right)+T\left(b\right)$
$T\left(ca\right)=cT\left(a\right)$
$T\left(0\right)=0$
Example:
A. $T\left({x}_{1},{x}_{2},{x}_{3}\right)=\left({x}_{1},0.{x}_{3}\right)$
$T\left({x}_{1}+{y}_{1},{x}_{2}+{y}_{2},{x}_{3}+{y}_{3}\right)=\left({x}_{1}+{y}_{1},0\left({x}_{2}+{y}_{2}\right),{x}_{3}+{y}_{3}\right)=T\left({x}_{1},0,{x}_{3}\right)+T\left({y}_{1},0,{y}_{2}\right)$
$T\left(c{x}_{1},c{x}_{2},c{x}_{3}\right)=T\left(c{x}_{1},\left(c\right)0,c{x}_{3}\right)=cT\left({x}_{1},0,{x}_{3}\right)$
$T\left(0,0,0\right)=0$
B. $T\left({x}_{1},{x}_{2}\right)=\left(2{x}_{1}-3{x}_{2},{x}_{1}+4,5{x}_{2}\right)$
$T\left({x}_{1}+{y}_{1},{x}_{2}+{y}_{2}\right)=\left(2\left({x}_{1}+{y}_{1}\right)-3\left({x}_{2}+{y}_{2}\right),\left({x}_{1}+{y}_{1}\right)+4,5\left({x}_{2}+{y}_{2}\right)\right)=\left(2{x}_{1}+2{y}_{1}-3{x}_{2}-3{y}_{2},{x}_{1}+{y}_{1}+4,5{x}_{2}+5{y}_{2}\right)$

Expert

For T to be a linear transformation, two criteria need to be satisfied:
1. $T\left(x+y\right)=T\left(x\right)+T\left(y\right)$
$T\left(ax\right)=aT\left(x\right)$ for a a scalar/constant.
As an example, suppose . Lets

nick1337

Expert

$T\left(a\stackrel{\to }{x}\right)=T\left(a{x}_{1},a{x}_{2},a{x}_{3}\right)=\left(a{x}_{1},0,a{x}_{3}\right)=a\left({x}_{1},0,{x}_{3}\right)=aT\left(\stackrel{\to }{x}\right)$

$T\left(\stackrel{\to }{x}+\stackrel{\to }{y}\right)=T\left({x}_{1}+{y}_{1},{x}_{2}+{y}_{2},{x}_{3}+{y}_{3}\right)=\left({x}_{1}+{y}_{1},0,{x}_{3}+{y}_{3}\right)=\left({x}_{1},0,{x}_{3}\right)+\left({y}_{1},0,{y}_{3}\right)=T\left(\stackrel{\to }{x}\right)+T\left(\stackrel{\to }{y}\right)$

Therefore the first transformation is linear as you correctly guessed. Just repeat the same procedure for B-E and see if it works or not.

Do you have a similar question?