Linear transformation from $\mathbb{R}^3 \rightarrow P^{2}[x]$ I'm pretty new to linear algebra, so I'd really appreciate your help with my problem.

Find the kernel and image for the given linear transformation: $A:\mathbb{R}^3 \to P_2[x]$
$A(1,0,-1)=x^2$
  $A(2,1,0)=x^2-x$
  $A(0,2,0)=3x$  

 A: The firt step is to find the standard matrix for the linear transformation.$$A(0,1,0)=A(0,2,0)/2=\frac{3x}{2}$$
$$A(1,0,0)=(A(2,1,0)-A(0,1,0))/2=(x^2-x-\frac{3x}{2})/2=\frac{x^2}{2}-\frac{5x}{4}$$
$$A(0,0,1)=-(A(1,0,-1)-A(1,0,0))=-(x^2-\frac{x^2}{2}+\frac{5x}{4})=-\frac{x^2}{2}-\frac{5x}{4}$$
Therefore$$A=\begin{bmatrix}\frac12&-\frac54&0\\0&\frac32&0\\-\frac12&-\frac54&0\end{bmatrix}$$The kernel of a transformation is vector that makes the transformation equal to the zero vector.
Create a system of equations:$$\left[\begin{array}{ccc|c}\frac12&-\frac54&0&0\\0&\frac32&0&0\\-\frac12&-\frac54&0&0\end{array}\right]$$
It is easy to get that $v_1=0,v_2=0,v_3$ is free.
The kernel is therefore spanned by $\begin{bmatrix}0\\0\\1\end{bmatrix}$
A: As @hardmath suggests, this problem can be solved by inspection once you’ve verified that the three given vectors of $\mathbb R^3$ indeed form a basis for that space. The image of the transformation is the span of $\{x^2, x^2-x, 3x\}$, which is easily seen to be the same as the span of $\{x,x^2\}$. By the rank-nullity theorem, this means that the kernel is one-dimensional, so we can find it by finding a linear combination of the three basis vectors that is mapped to zero. Working backwards from the image, we can quickly discover that one such combination is $$(x^2)-(x^2-x)-\frac13(3x)=A[(1,0,-1)-(2,1,0)-\frac13(0,2,0)]=A\left(-1,-\frac53,-1\right)$$ therefore the kernel of $A$ is spanned by $\left(-1,-\frac53,-1\right)$ or, more simply, by $(3,5,3)$.  
A more systematic approach is to compute the matrix of the transformation, then apply row-reduction to find the kernel and image. We can immediately write down the matrix relative to the given basis $B$—its columns are the images of the three basis vectors: $$[A]_B=\begin{bmatrix}0&0&0\\0&-1&3\\1&1&0\end{bmatrix}$$ which row-reduces to $$\begin{bmatrix}1&0&3\\0&1&-3\\0&0&0\end{bmatrix}.$$ From this we can see that the first and second columns of $[A]_B$ form a basis for the image, i.e., the image is the span of $\{x^2,x^2-x\}$, which is clearly the same space as found above. The row-reduced matrix also tells us that the kernel is spanned by $(-3,3,1)$, but this is relative to the basis $B$. Remembering that a coordinate vector stands for a linear combination of the basis vectors, we find that the kernel is spanned by $-3\cdot(1,0,-1)+3\cdot(2,1,0)+1\cdot(0,2,0)=(3,5,3)$. We could also have found a basis for the image by column-reducing $[A]_B$: the non-zero columns of the resulting matrix form a basis for the image. I prefer doing it that way because it usually results in a more useful basis.  
Another way to proceed is to work with the matrix of the transformation relative to the standard basis $S$ instead. One way to find this matrix is to perform a change of basis on $[A]_B$: $$[A]_S = \begin{bmatrix}0&0&0\\0&-1&3\\1&1&0\end{bmatrix}\begin{bmatrix}1&2&0\\0&1&2\\-1&0&0\end{bmatrix}^{-1} = \begin{bmatrix}0&0&0\\0&-1&3\\1&1&0\end{bmatrix}\begin{bmatrix}0&0&-1\\\frac12&0&\frac12\\-\frac14&\frac12&-\frac14\end{bmatrix} = \begin{bmatrix}0&0&0\\-\frac54&\frac32&-\frac54\\\frac12&0&-\frac12\end{bmatrix}.$$ (Note that the change-of-basis matrix has the elements of $B$ as its columns.) This time, row-reduction yields $$\begin{bmatrix}1&0&-1\\0&1&-\frac53\\0&0&0\end{bmatrix}.$$ We can read the kernel directly from this matrix: it is spanned by $\left(1,\frac53,1\right)$. The image basis is a bit messier this time, though. It’s the span of $\frac12x^2-\frac54x$ and $\frac32x$, which is of course the same space as before.  
[N.B.: To reduce clutter, I’ve written all of the vectors as row vectors above. They are all really column vectors, of course.]
