What is the dimension of image $T(p)(x)=p(x+1)-p(x)$? Let $V$ be the vector space of all polynomials $p(x)$, $x \in \mathbb{R}$, with real coefficients of degree at most 3. Consider the linear map $T: V \to V$ given by $T(p)(x)=p(x+1)-p(x)$. Then, the dimension of the image $T(V)$ is $?$
Attempt: Consider an arbitrary polynomial $ax^3+bx^2+cx+d$. Now,  $p(x+1)-p(x)= 3ax^2+(3a+2b)x+(a+b+c)$
Arranging the coefficients in a matrix, 
\begin{pmatrix}
3&0&0\\3&2&0\\1&1&1
\end{pmatrix}which has a rank of $3$. Hence, $\dim(Im(T))=3$
. Is this alright?
 A: Your solution is correct. One can also solve it this way: assume $p(x)\in\ker T$. Then $T(p(x))=0$ and after equating the coefficients you already found to $0$ we get the system:
$$
\begin{cases}
3a=0 \\
3a+2b=0 \\
a+b+c=0
\end{cases}
$$
Clearly, the only solution to this system is $(a, b, c)=(0, 0, 0)$ and so $d$ is a free variable, so $\ker T=\operatorname{Span}\{1\}$ (all constant polynomials), so $\dim\ker T=1$ and $\dim\operatorname{Im}T=4-\dim\ker T=3$ (rank-nullity theorem)
A: The answer is correct, and your reasoning is almost there.
First of all, the dimension of $V$ is $4$, so you should have a $4 \times 4$ matrix to represent the transformation. 
It sounds like you want to take $\beta = \{x^3, x^2, x, 1\}$ as an ordered basis for $V$. Then, by your computation, the matrix for $T$ with respect to $\beta$ should take $(1,0,0,0) \mapsto (0,3,3,1)$, $(0,1,0,0) \mapsto (0,0,2,1)$, $(0,0,1,0) \mapsto (0,0,0,1)$, and $(0,0,0,1) \mapsto (0,0,0,0)$ (do you see why?)
Now, make a matrix out of this.
A: With the standard basis on $P_3$, $$B= \{1,x,x^2,x^3\}$$ The matrix of your transformation is $$\begin{pmatrix} 0&1&1&1\\0&0&2&3\\0&0&0&-3\\0&0&0&0\end{pmatrix}$$
With rank of $3$.
Thus the image of your transformation is three dimensional.
A: Here $\{1,x,x^2,x^3\}$  is a basis, so $$\text{Range }\; T=\text{span}\{ T(1),T(x),T(x^2),T(x^3)\}=\text{span}\{ 0,1,1+2x,3x^2+3x+1\}$$   From the above, we conclude, the dimension of Range $T$ is three
