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A linear transfomation T:$\mathbb{P}_{3}\rightarrow\mathbb{P}_{3}$such that T{[}p(x){]}=p''(x) +p(x).Check

  1. T is one-one?

2.T is onto?

MY APPROACH: Ker(T) is given by p''(x)+p(x)=0 a differential eqn whose solution is p(x)=Acosx +Bsinx.I think p(x) is periodic function hence Ker(T) is not Trivial.So T is neither one one nor onto.

ANOTHER APPROACH : T(1),T(x),T(x$^{2}$ ),T(x$^{3})$are Linearly independent vectors belonging to Range(T).Hence Ker(T) is trivial.Hence T is one one and onto.

  • 1
    But is your $p(x)$ in the domain of $T$? – 2017-02-16
  • 1
    The second approach is correct. The first one makes no sense as trigonometric functions are not in $P$ – 2017-02-16

1 Answers 1

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Your first approach without differential equations: let $p \in ker(T)$.

Then $p''+p=0$. Hence $p^{(4)}+p''=0$. Since $p^{(4)}=0$, we get $p''=0$ and therefore $p=0$.