$T: V \to \Bbb{R^2}$ by $T(f)=(f'(0),f(0))$. Let V be the space of twice differentiable function on $\Bbb{R}$ such that $$f''-2f'+f=0.$$
Define $T: V \to \Bbb{R^2}$ by $$T(f)=(f'(0),f(0)).$$
The I could see that $T$ is one-one, but is $T$ onto?
How can I prove/disprove it?
 A: This is a classic existence and uniqueness problem. You know that solutions to second order initial value problems all exist, and are unique, so $T$ is onto because for every $a,b$ there exists a solution $f$ to the given O.D.E. satisfying $f'(0)=a$ and $f(0)=b$. One-to-one follows from the uniqueness of the solution.
A: Any linear transformation between finite-dimensional vector spaces of the same dimension is injective if and only if it is surjective by the rank-nullity theorem.
However, it is not obvious to me that our $T$ is, in fact, injective. So, to solve the problem let's compute a matrix representation of $T$. First, we must provide a basis $\alpha$ for $V$. To do so, note that the equation defining $V$ is a second order ODE so the theory of ODE's tells us that $\dim V=2$. Moreover, the characteristic polynomial of the ODE is 
$$
\chi(t)=t^2-2\,t+1=(t-1)^2
$$
This tells us that $f_1,f_2\in V$ where 
\begin{align*}
f_1(t) &= e^t & f_2(t) &= te^t
\end{align*}
Moreover, $f$ and $g$ are linearly independent since the Wronskian
$$
W(t)=\det\begin{bmatrix}f_1(t) & f_2(t) \\ f^\prime(t) & g^\prime(t) \end{bmatrix}=\det\begin{bmatrix} e^t & te^t\\ e^t& (t+1)e^t\end{bmatrix}=e^{2\,t}
$$
satisfies $W(0)\neq 0$ (a collection of $n$ smooth functions is linearly independent if their Wronskian does not vanish identically). Since $\dim V=2$ this implies that $\alpha=\{f_1,f_2\}$ is a basis for $V$.
Of course, we also have the standard basis $\beta=\{\vec e_1,\vec e_2\}$ for $\Bbb R^2$.
Now, to compute the matrix $[T]_\alpha^\beta$ we compute
\begin{array}{rcrcrcrcrcrc}
T(f_1) &=& \color{red}{1}\, \vec e_1 &+& \color{blue}{1}\, \vec e_2 \\
T(f_2) &=& \color{red}{0}\, \vec e_1 &+& \color{blue}{1}\, \vec e_2
\end{array}
This implies
$$
[T]_\alpha^\beta=\begin{bmatrix}\color{red}{1} & \color{red}{0} \\ \color{blue}{1}&\color{blue}{1}\end{bmatrix}
$$
Of course, this matrix is invertible so $T$ is invertible.
A: What is the dimension of $V$ ? $V$ is a two dimensional real vector space. So the By the rank nullity theorem, $dim_{\mathbb{R}} V= dim_{\mathbb{R}} ker T + dim_{\mathbb{R}} Im T$. So $T$ is onto if and only if $T$ is one to one.
A: If you know some basics about linear ODE, then you could say that a basis for the solutions is given by $e^x, xe^x$, so the solution space is two-dimensional.
Thus, since injectivity implies surjectivity (by rank-nullity), you're finished.
Or, if you want, you could explicitly write out the matrix for the linear transformation in terms of the two natural bases and see that it has an inverse.
A: To be very explicit: the functions $e^x$ and $xe^x$ form a basis for $V;$ the map $T:V \to \mathbb{R}^2$ has the matrix representation $\begin{pmatrix} 1&1\\ 1&0 \end{pmatrix},$ which has full rank.
