Question 3.3: Determine whether the linear transformation T:P2 → R³, where......

Let T\big(a+b x+c x^{2}\big)=0

\therefore\!\!\left[\begin{array}{c c}{{2a-b}}\\ {{a+b-3c}}\\ {{c-a}}\end{array}\right]=\left[\ \begin{array}{c}{{0}}\\ {{0}}\\ {{0}}\end{array}\ \right].

Comparing both sides,

\therefore2a-b=0,\ \ a+b-3c=0,\ \ c-a=0.

Solving the above equations by Gauss elimination, we get

Augmented matrix = {\left[\begin{array}{c c c | c} &&&\!\! 0 \\[-0.5 em] {2}&{-1}&{0} & \\ {1}&{1}&{-3} & \!\! 0 \\ -1&0&1& \\[-0.5 em] &&&\!\!0 \end{array}\!\!\right]}.

Take R_{2}\leftrightarrow R_{1}

\approx {\left[\begin{array}{c c c | c} &&&\!\! 0 \\[-0.5 em] {1}&{1}&{-3} & \\ {2}&{-1}&{0} & \!\! 0 \\ -1&0&1& \\[-0.5 em] &&&\!\!0 \end{array}\!\!\right]}.

By taking R_{2}\to R_{2}-2R_{1},\;R_{3}\to R_{3}+R_{1}

\approx {\left[\begin{array}{c c c | c} &&&\!\! 0 \\[-0.5 em] {1}&{1}&{-3} & \\ {0}&{-3}&{\ \ 6} & \!\! 0 \\ 0&1&-2& \\[-0.5 em] &&&\!\!0 \end{array}\!\!\right]}

{\frac{-1}{3}}R_{2}

\approx {\left[\begin{array}{c c c | c} &&&\!\! 0 \\[-0.5 em] {1}&{1}&{-3} & \\ {0}&{1}&{-2} & \!\! 0 \\ 0&1&-2& \\[-0.5 em] &&&\!\!0 \end{array}\!\!\right]}

R_{3}\to R_{3}-R_{2}

\approx {\left[\begin{array}{c c c | c} &&&\!\! 0 \\[-0.5 em] {1}&{1}&{-3} & \\ {0}&{1}&{-2} & \!\! 0 \\ 0&0&0 \\[-0.5 em] &&&\!\!0 \end{array}\!\!\right]}.

By taking substitution

a+b-3c=0,\ \ b-2c=0,\ \ c=t\in R

a=t,b=2t,c=t\in R.

Solution set is S=\big\{(a,b,c){=}t(1,2,1)/t\in R\big\}.

The solution is trivial; hence, \ker(T)=\big\{(a,b,c)=t(1,2,1)/t\in R\big\}\neq\left\{0\right\}.

Hence, T is not one-one.

In the next section, we will discuss isomorphism and its example.