Determine whether the linear transformation T:R² → R², where T(x,y)=(x+y,x-y), is one-one, onto, or both or neither.

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Accepted Answer

A linear transformation is one-one if and only if ker(T)={0}[latex]\ \quad[/latex]Let

[latex]\quad\quad\quad\quad\quad\quad\quad \begin{array}{l}{{T(x,y)=\overline{{{0}}}}}\ {{ herefore\left(x+y,x-y ight)=(0,0)}}\ {{x+y=0,x-y=0.}}\end{array}[/latex]

Solving these equations,

[latex]\ \quad\quad\quad\quad\quad\quad\quad\quad x=0,\ \ y=0.[/latex]

The solution is trivial; hence, ker(T)={0}.[latex]\ \quad[/latex] Hence, T is one-one. [latex]\ \quad[/latex] A linear transformation is onto if R(T)=W.[latex]\ \quad [/latex] Let [latex] u=\left(x,y ight)[/latex] and [latex]w=(a,b)[/latex] be in R², where a and b are the real numbers such that [latex]T(V)=W.[/latex]

[latex]\quad\quad\quad\quad\quad\quad\quad \begin{array}{l}{{T\left(x,y ight)=\left(a,b ight)}}\ {{\because \left(x+y,x-y ight)=\left(a,b ight)}}\ {\because x+y=a,x-y=b.}\end{array}[/latex]

Solving these equations,

[latex]\quad\quad\quad\quad\quad\quad\quad x={\frac{a+b}{2}},\ \ y={\frac{a-b}{2}}.[/latex]

Thus, for every [latex]w=(a,b)[/latex] in R², there exists a [latex] u=\left({\frac{a+b}{2}},{\frac{a-b}{2}} ight)[/latex] in R².

[latex]\quad [/latex] Hence, T is onto.

Question 3.1: Determine whether the linear transformation T:R² → R², where......

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