Anish Buchanan

2021-01-06

Use a counterexample to show that the statement is false.

$T:{R}^{2}\to {R}^{2},T({x}_{2},{x}_{2})=({x}_{1}+4,{x}_{2})$ is a linear transformation?

oppturf

Skilled2021-01-07Added 94 answers

Approach:

Let V and W be vector spaces. The function$T:V\to W$ is a linear transformation of V into W when the two properties below are true for all u and v in V and for any scalar c.

$T(u+v)=T\left(u\right)+T\left(v\right)$

$T(cu)=cT(u)$

Calculation:

Assume the two vectors$u=({u}_{1},{u}_{2}){\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}v=({v}_{1},{v}_{2}).$

The sum of the two vectors is,

$u+v=({u}_{1},{u}_{2})+({v}_{1},{v}_{2})$

$=({u}_{1}+{v}_{1},{u}_{2}+{v}_{2})$

Apply transformation on both side of the above equation.

$T(u+v)=T({u}_{1}+{v}_{1},{u}_{2}+{v}_{2})$

$=({u}_{1}+{v}_{1}+4,{u}_{2}+{v}_{2})\dots \left(1\right)$

The sum of the two transformations is,

$T\left(u\right)+T\left(v\right)=T({u}_{1},{u}_{2})+T({v}_{1},{v}_{2})$

$=({u}_{1}+4,{u}_{2})+({v}_{1}+4,{v}_{2})$

$=({u}_{1}+{v}_{1}+8,{u}_{2}+{v}_{2})\dots \left(2\right)$

From equation (1) and equation (2),

$T(u+v)\ne T\left(u\right)+T\left(v\right)$

From the above result, the function$T({x}_{1},{x}_{2})=({x}_{1}+4,{x}_{2})$ is not a linear transformation.

Therefore, the statement is false.

Let V and W be vector spaces. The function

Calculation:

Assume the two vectors

The sum of the two vectors is,

Apply transformation on both side of the above equation.

The sum of the two transformations is,

From equation (1) and equation (2),

From the above result, the function

Therefore, the statement is false.

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