$\displaystyle T(a{x^2} + bx + c) = \left( {\begin{array}{*{20}{c}} {p(0)}&{p'(0)}\\ {p'(0)}&{p''(0)} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} c&b\\ b&{2a} \end{array}} \right)$
a) It should be easy to show $T(\alpha p+q)=\alpha T(p)+T(q)$
b) Looking at the mapping what polynomial(s) map to $\displaystyle \left( {\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}} \right)~?$
c) for injectivity If T(p)=T(q) does it follow that $p=q$ HINT: what if $\displaystyle \left( {\begin{array}{*{20}{c}} c&b\\ b&2a \end{array}} \right)=\left( {\begin{array}{*{20}{c}} f&e\\ e&2d \end{array}} \right)~?$
For surjectivity what happens in the case of $\displaystyle \left( {\begin{array}{*{20}{c}} 0&2\\ 1&3 \end{array}} \right)~?$