Given Information
We are given with a transformation \[ T : P _ { 2 } \rightarrow R ^ { 2 } \text { by } T ( \mathbf { p } ) = \left[ \begin{array} { l } { \mathbf { p } ( 0 ) } \\ { \mathbf { p } ( 1 ) } \end{array} \right] \] We have to prove that the transformation is linear. We also have to find a polynomial that spans kernel of T.

Step-1: (a)
Let us take two polynomials. \[ \begin{array} { l } { \mathbf { p } ( t ) = a _ { 1 } + b _ { 1 } t + c _ { 1 } t ^ { 2 } } \\ { \mathbf { q } ( t ) = a _ { 2 } + b _ { 2 } t + c _ { 2 } t ^ { 2 } } \end{array} \]

Step-2: Check for vector addition
\[\begin{array}{l} T({\bf{p}} + {\bf{q}}) = \left[ {\begin{array}{*{20}{l}} {({\bf{p}} + {\bf{q}})(0)}\\ {({\bf{p}} + {\bf{q}})(1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {\left( {{a_1} + {a_2}} \right) + \left( {{b_1} + {b_2}} \right)(0) + \left( {{c_1} + {c_2}} \right){{(0)}^2}}\\ {\left( {{a_1} + {a_2}} \right) + \left( {{b_1} + {b_2}} \right)(1) + \left( {{c_1} + {c_2}} \right){{(1)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a_1} + {a_2}}\\ {\left( {{a_1} + {a_2}} \right) + \left( {{b_1} + {b_2}} \right) + \left( {{c_1} + {c_2}} \right)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a_1} + {a_2}}\\ {{a_1} + {a_2} + {b_1} + {b_2} + {c_1} + {c_2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a_1} + {a_2}}\\ {\left( {{a_1} + {b_1} + {c_1}} \right) + \left( {{a_2} + {b_2} + {c_2}} \right)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a_1}}\\ {{a_1} + {b_1} + {c_1}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {{a_2}}\\ {{a_2} + {b_2} + {c_2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a_1}}\\ {{a_1} + {b_1} + {c_1}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {{a_2}}\\ {{a_2} + {b_2} + {c_2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a_1} + {b_1}(0) + {c_1}{{(0)}^2}}\\ {{a_1} + {b_1}(1) + {c_1}{{(1)}^2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {{a_2} + {b_2}(0) + {c_2}{{(0)}^2}}\\ {{a_2} + {b_2}(1) + {c_2}{{(1)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {{\bf{p}}(0)}\\ {{\bf{p}}(1)} \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} {{\bf{q}}(0)}\\ {{\bf{q}}(1)} \end{array}} \right]\\ = T({\bf{p}}) + T({\bf{q}}) \end{array}\] Hence the transformation is closed under vector addition.

Step-3: Check for scalar multiplication
Compute transformation of $cp$ \[\begin{array}{l} T(c{\bf{p}}) = \left[ {\begin{array}{*{20}{c}} {(c{\bf{p}})(0)}\\ {(c{\bf{p}})(1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {c{\bf{p}}(0)}\\ {c{\bf{p}}(1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {c{a_1} + c{b_1}(0) + c{c_1}{{(0)}^2}}\\ {c{a_1} + c{b_1}(1) + c{c_1}{{(1)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {c{a_1}}\\ {c{a_1} + c{b_1} + c{c_1}} \end{array}} \right]\\ = c\left[ {\begin{array}{*{20}{r}} {{a_1}}\\ {{a_1} + {b_1} + {c_1}} \end{array}} \right]\\ = c\left[ {\begin{array}{*{20}{l}} {{a_1} + {b_1}(0) + {c_1}{{(0)}^2}}\\ {{a_1} + {b_1}(1) + {c_1}{{(1)}^2}} \end{array}} \right]\\ = c\left[ {\begin{array}{*{20}{l}} {{\bf{p}}(0)}\\ {{\bf{p}}(1)} \end{array}} \right]\\ = cT({\bf{p}}) \end{array}\] Therefore, the transformation is closed under scalar multiplication. Therefore,

The transformation is a linear transformation

Step-4: (b) The polynomial that spans the kernel
For kernel of T is all polynomials such that $T\left( p \right) = 0$. Let the polynomial is: \[ \mathbf { p } ( t ) = a + b t + c t ^ { 2 } \] Apply the above conditions to polynomial. \[{\bf{p}}(0) = 0 \Rightarrow a + b(0) + c{(0)^2} = 0 \Rightarrow a = 0\] \[\begin{array}{l} {\bf{p}}(1) = 0 \Rightarrow a + b(1) + c{(1)^2} = 0\\ a + b + c = 0\\ 0 + b + c = 0\\ c = - b \end{array}\] Therefore the kernel of T can have the form \[ \begin{aligned} \mathbf { p } ( t ) & = b \left( t - t ^ { 2 } \right) \\ & = b t ( 1 - t ) \end{aligned} \]

Step-5: (b) Range of p
The range of $T$ consists of all possible vectors in $R^2$ that satisfy $p(0)=0$ and $p(1)=0$. For a general form of $p\left( t \right) = a + bt + c{t^2}$. So, by applying the condition \[\begin{array}{l} p\left( 0 \right) = a;\,\,\,\\ p\left( 1 \right) = a + b + c \end{array}\] So, the range of T is all values in $R^2$