Q:

Does A linear Transformation Of This Kind Exist?

Is there a linear transformation $T:\mathbb{R}_2[x] \rightarrow \mathbb{R}^3$
So that
$Ker(T)=span \{ 1+x-x^2, 2+3x^2 \}$

$Im(T)=span \{ (0,0,9) \}$
What I have done:
$Ker(T)=(\alpha+\alpha x+\alpha x^2+2\beta+3\beta x^2)=(\alpha+2\beta)+\alpha x+(3\beta-\alpha)x^2$
by definition the expression $(\alpha+2\beta)+\alpha x+(3\beta-\alpha)x^2=0$
and it can be only if $\alpha=\beta=0$ therefore $Ker(T)={0}$ and $Dim(Ker(T))=0$.
By $DimV=DimKer(T)+DimIm(T)$ the dimension of Im(T) is  3=0+Im(T) but Dim(Im(T))=1 and therefore there is no such linear transformation
Is this poof valid?  

A:

As a vectors pace, $\mathbb{R}_2[x] \cong \mathbb{R}^3$, so we can reduce the problem to there.
Consider the general question of whether there is a linear map $T : \mathbb{R}^3 \to \mathbb{R}^3$ such that the kernel is a specific two-dimensional subspace of $\mathbb{R}^3$ and the image of a specific one-dimensional subspace of $\mathbb{R}^3$.
I claim that the solution to any one problem of this form gives a solution to all problems of this form. By a change of basis, all "two-dimensional kernel one-dimensional image" problems are really the same problem.
(try figuring it out on your own before reading on)
By a change of basis -- i.e. an invertible linear transformation on $\mathbb{R}^3$ -- we can always convert any $m$-dimensional subspace into any other $m$-dimensional subspace. We need a different change of basis transformation for each one, but that's fine.
Suppose $S$ is any matrix with a two-dimensional kernel and one-dimensional image. Let $B_0$ be any change of basis that maps the space we want to be $\ker(T)$ to $\ker(S)$, and let $B_1$ be one that maps $im(T)$ to $im(S)$.
Putting all of this data into a diagram:
$$\begin{matrix}\mathbb{R}^3 &\xrightarrow{T}& \mathbb{R}^3
\\\!\!\!\!\!\!\!\!\!\!B_0 \downarrow& & \downarrow B_1 \!\!\!\!\!\!\!\!\!\!
\\\mathbb{R}^3 &\xrightarrow{S}& \mathbb{R}^3
\\\end{matrix}$$
That is, $B_1 T = S B_0$, and thus $T = B_1^{-1} S B_0$ will have the desired spaces as kernel and image.
So if you can solve any instance of the "two-dimensional kernel one-dimensional image" problem and find the two needed change-of-basis transformations, you can find $T$.

