Kernel, basis and image Let$P_n(R)$ denote the vector space of real polynomial functions of degree less than or
equal to $n$, let $p_i$ denote the polynomial determined by $p_i(x) =x^i$, and let 
$F:P_2(R)\rightarrow P_3(R)$.
be the linear transformation determined by $$F(f)(x) =\int^{x+1}_{2-x}(1-t)f(t) dt:$$
Determine a basis for the kernel of F. Determine a basis for the image of F. Define 
$A :=[p_0,p_1,p_2,p_3]$ and compute $M^A_B(F)$, the matrix of F with respect to the ordered bases B and A. Compute $M^A_C(F)$ and give the rank of $M^A_C(F)$. 
 A: Using a representing matrix is definitely the way to go. Consider on the codomain the basis consisting of $q_0=1$, $q_1=x$, $q_2=x^2$, $q_3=x^3$ (in order to avoid confusion I'll distinguish between vectors in $P_2$ and vectors in $P_3$).
We have just to compute, with some patience, $F(p_0)$, $F(p_1)$ and $F(p_2)$; I won't make the computation, but just show you the way.
You'll find $F(p_i)=a_iq_0+b_iq_1+c_iq_2+d_iq_3$ (for $i=0,1,2$); write the three sets of coefficients to form columns and put them together, getting
$$\begin{bmatrix}
a_0 & a_1 & a_2 \\
b_0 & b_1 & b_2 \\
c_0 & c_1 & c_2 \\
d_0 & d_1 & d_2
\end{bmatrix}=\begin{bmatrix} v_1 & v_2 & v_3\end{bmatrix}$$
which is the matrix representing $F$ with respect to the bases $\{p_0,p_1,p_2\}$ and $\{q_0,q_1,q_2,q_3\}$.
Now do Gaussian elimination on this matrix. The columns with the pivots tell you what vectors to choose among $F(p_0)$, $F(p_1)$ and $F(p_2)$ to form a basis of the image. Let's say they are the first and the second (I really don't know what they are, nor if the rank is $2$): then $\{F(p_0),F(p_1)\}$ is a basis of the image.
In this case you'll also have a linear relation telling you that $v_3=\alpha v_1+\beta v_2$, so you know that $F(p_2)=h F(p_0)+k F(p_1)$ and so the representing matrix with respect to $\{p_0,p_1,p_2\}$ of the linear map, though as having the image as codomain is
$$
\begin{bmatrix}
1 & 0 & h\\
0 & 1 & k
\end{bmatrix}
$$
So, under these hypotheses, then you already have a basis for the kernel, because you know that
$$
F(hp_0+kp_1-p_2)=0
$$
so the basis consists of the single polynomial $h+kx-x^2$.
It would be similar if you find that the pivot columns are the first and the third; in this case you know that $F(p_1)=hF(p_0)$, so $\{F(p_0),F(p_2)\}$ is a basis of the image. For the kernel, you'd find $F(hp_0-p_1)=0$.
If the rank is $1$ the things are just a bit more complicated.
A: Let's check some ideas:
Since $\;f(t)\in P_2(\Bbb R)\;$ , we can write $\;f(t)=at^2+bt+c\;,\;\;a,b,c\in\Bbb R\;$:
$$f(t)\in\ker F\iff 0=F(f(t)):=\int\limits_{2-x}^{x+1}(1-t)(at^2+bt+c)dt=$$
$$\left.\left[-\frac a4t^4+\frac{a-b}3t^3+\frac{b-c}2t^2+ct\right]\right|_{2-x}^{x+1}=$$
$$=-\frac a4\left((x+1)^4-(2-x)^4\right)+\frac{a-b}3\left((x+1)^3-(2-x)^3\right)+\frac{b-c}2\left((x+1)^2-(2-x)^2\right)+c\left(x+1-(2-x)\right)=$$
$$=-\frac a4\left(12x^3-18x^2+36x+16\right)+\ldots$$
From the above, we already have that $\;a=0\;$ (look at the coefficient of $\;x^3\;$ from the first parentheses, which does not appear again in the whole expression...)
Well, find now all the conditions on the coefficients of $\;f\;$ needed to have $\;f\in\ker F\;$ (yes, this looks disgustingly ugly and annoying...too bad)
As for $\;M_A^B(F)\;$ and etc.: you didn't give any $\;B, C\;$, only $\;A\;$ but who cares? You have to apply $\;F\;$ on $\;A\;$ and write the result as a linear combination of $\;B\;$ and then take the transpose of the coefficients matrix.
I have to say that unless there is a slick trick that I'm missing to do this, this is one of the most disgustingly annoying and calculation-hard exercises in basic linear algebra I've ever had the unpleasure to meet...My deepest condolences to the OP, yet the exercise is basic.
