Linear transformation defined on vector space of twice differentiable functions to $R^2$.How to determine whether it is one to one or onto

Question

Given that T is a linear transformation from V to $R^2$ where V be the vector space of all twice differentiable functions with the condition that $f''(0)-2f'(0)+f(0)=0$ and $T(f)=(f'(0),f(0))$ Then how to determine whether it is one one or onto I was trying to find out the dimension of the vector space V
But I am unable to find an basis of V How to proceed with this please help

Answer 1

It is not injective, consider a función whose Taylor series around zero has a not zero term of degree 3. But it is surjective, you can take any pair of numbers and construct the corresponding polynomial.

Answer 2

$T$ is not one-one because if $f$ is any polynomial of the form $x^2a(x)$, where $a(x)$ is some polynomial, then $f \in V$ and and $f'(x)=x^2a'(x)+2xa(x)$. Thus $$T(f)=(0,0) \qquad \text{for all such } f(x).$$

$T$ is onto because if we take any $(a,b) \in \Bbb{R}^2$, then we can have $f(x)=\left(\frac{2a-b}{2}\right)x^2+ax+b \in V$ such that $T(f(x))=(a,b)$ because $f(0)=b$ and $f'(0)=a$.

For the dimension of $V$, note that $\{x^3,x^4, \ldots\} \subset V$ as all $f(x)=x^n$ (with $n \geq 3$) satisfy the given condition $f''(0)-2f'(0)+f(0)=0$. Furthermore the set is linearly independent. Thus $V$ is not finite dimensional.