I have serious doubts - I will be very grateful if someone will help me here
Investigate whether the given transformation is a monomorphism / epimorphism. Find its image and kernel. $$ F \in L(\mathbb R[x]_3,\mathbb R[x]_3), F(p)(t) = p(t+1) - p(t) $$
My try
Okay, let
$$ p(t) = ax^3 + bx^2 + cx + d $$
then
$$F(p)(t) = ... = 3ax^2 + 3ax + 2bx + 3 = x^2(3a)+x(3a+2b) +3 = a(3x^2+3x) + b(2x) + 3 $$
Is it monomorphism?
let $m_0 = 0 \wedge m_1 = 3a \wedge m_2 = 3a+b \wedge m_3 = 3$
so $a = \frac{m_1}{3} \wedge b = \frac{m_2-m_1}{2} $
so $a$ and $b$ are determined unambiguously so it is monomorphism
It is not epimorphism because $ x^3 \in\mathbb R[x]_3 $ but I can't get $x^3$ in use of $F$
$\ker F = \left\{ 0 \right\} $ because of part $ 3 $ it will be never polynomial zero
If it comes to image, we checked that: $$F(p)(t) = ... = a(3x^2+3x) + b(2x) + 3 $$ so $ im(F) = span(3x^2+3,2x,3) $ Moreover this system is lineary independent so $rank(F) = 3$
Have I done this correctly? Or there is something to fix?
You have a contradiction in your solution: if the map is injective it must also be surjective, because it is a linear map of a finite dimensional vector space to itself.
It's much easier with matrices. If you consider the standard basis $\{p_0=1,p_1=t,p_2=t^2,p_3=t^3\}$, then you see that $$ F(p_0)=0,\quad F(p_1)=1,\quad F(p_2)=2t+1\quad F(p_3)=3t^2+3t+1 $$ and therefore the matrix of $F$ is \begin{bmatrix} 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}
What's the rank of this matrix?
Where did you go wrong? The image of $p(t)=at^3+bt^2+ct+d$ is \begin{align} p(t+1)-p(t) &=at^3+3at^2+3at+a+bt^2+2bt+b+ct+d-at^3-bt^2-ct-d\\ &=3at^2+(3a+2b)t+(a+b+c) \end{align} which is zero when \begin{cases} 3a=0\\ 3a+2b=0\\ a+b+c=0 \end{cases} so when $a=b=c=0$. But $d$ can be anything. Thus the nullity is $1$ and the rank is $3$.