$T:P_2(x) \to P_3(x)$, where $T(p(x))=xp(x)$
So T can be written as $T(c+bx+ax^2) = (cx + bx^2 + ax^3)$
I'm finding it hard to see the one-to-one nature here, although I am pretty sure I saw in an earlier chapter in the book that it is one-to-one. Maybe I'm looking for a new perspective on how you show this here.I think I'm supposed to be looking at the coefficients and not the x's, is that correct?
For surjectivity, well if I look at the coefficients, $cx + bx^2 + a x^3$ is actually $0d + cx + bx^2 + ax^3$ so only these vectors in the whole of $P_3$ space is being mapped too, so it is not onto.
Any help on injectivity?
In order to show that $T$ is injective, given $p$ and $q$ in $P_2(x)$, we have to show that $T(p) = T(q)$ implies $p=q $ (it's equivalent to $p\neq q$ imply $T(p) \neq T(q)$).
So, let $p=c_1+b_1x+a_1x^2$ and $q=c_2+b_2x+a_2x^2$ and suppose $$T(p) = T(q).$$ Then $$c_1x+b_1x^2+a_1x^3 = c_2x+b_2x^2+a_2x^3.$$ In other words (two polynomials are equal if they have the same coefficients) $$c_1=c_2, \quad b_1=b_2, \quad a_1=a_2.$$ It means that $p=q$.