Question 3.2: Determine whether the linear transformation T:R² → R³, where......
Determine whether the linear transformation T:R² → R³, where T(x,y)=(x-y,y-x,2x-2y), is one-one, onto, or both or neither.
A linear transformation is one-one if and only if ker(T)={0}\\ \quadLet
\begin{array}{l}{{T(x,y)=\overline{{{0}}}}}\\[1 em] {{\therefore\left(x-y,y-x,2x-2y\right)=(0,0,0)}}\\[1 em] {{x-y=0,\ y-x=0,\ 2x-2y=0.}}\end{array}
Solving these equations, x=y
x=t,\ y=t,\operatorname{where},t\in R.
Solution set is S=\Big\{\big(x,y\big)=t(1,1)/t\in R \Big\}.
The solution is trivial; hence, \ker(T)=\Big\{\big(x,y\big)=t(1,1)/t\in R\Big\}\neq \{0\}.
Hence, T is not one-one.
A linear transformation is onto if R(T)=W .
\qquadLet \nu=\left(x,y\right) in R² and w=(a,b,c) ∈ R³, where a,b and c are the real numbers such that T(V)=W
\begin{array}{l}{{T\left(x,y\right)=\left(a,b,c\right)}}\\[1 em] {{\because \left(x-y,y-x,2x-2y\right)=\left(a,b,c\right)}}\\[1 em] {\because x-y=a,\ y-x=b,\ 2x-2y=c.}\end{array}
Solving these equations,
the system has a solution only when a=-b=\frac{c}{2} not for all a, b and c.
Hence, T is not onto.