<s>[INST] Find the basis of the Ker(T) and Im(T) for $T: P_3 \to P_3$, $T(ax^3+bx^2+cx+d)=dx^3+cx^2+bx+a$ For the linear application, $T: P_3 \to P_3$, $T(ax^3+bx^2+cx+d)=dx^3+cx^2+bx+a$, find


*

*A matrix representation with respect to a basis of your choice

*A basis of Ker(T)

*A basis of Im(T)


What I have done so far...
Matrix representation: Taking a closer look at the map and the matrix operations, we can easily see that the matrix representation will be $\ldots$
Corrected Matrix:
$\begin{pmatrix} 0 & 0 & 0 & d \\ 0 & 0 & c & 0 \\ 0 & b & 0 & 0 \\ a & 0 & 0 & 0  \end{pmatrix}  $
Because 
$\begin{pmatrix} d & 0 & 0 & 0 \\ 0 & c & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & a  \end{pmatrix} \begin{pmatrix} x^3 \\ x^2 \\ x \\ 1 \end{pmatrix} =T(ax^3+bx^2+cx+d)$
Basis of Ker(T): You have to find the set of all $ax^3+bx^2+cx+d \in P_3$ such that $$T(ax^3+bx^2+cx+d)=(0,0,0,0)^T$$
We consider 
$\begin{pmatrix} d & 0 & 0 & 0 \\ 0 & c & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & a  \end{pmatrix} \begin{pmatrix} x^3 \\ x^2 \\ x \\ 1 \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$
Which will result in the following equations:
\begin{cases} 
      dx^3=0 \\
      cx^2=0 \\
      bx=0 \\
      a=0 \\
      \end{cases}
Is this correct? I do not know how to proceed. I know we have to find $ax^3+bx^2+cx+d=0$ in $P_3$ such that T(ax^3+bx^2+cx+d)=0.
Basis of Im(T): If we just want to have a set of vectors which span the image, then it would be enough to take the column-vectors. But we want to have linear-independent vectors, which span our image. 
What we have to do now: Consider $T^{T}$ matrix of T and apply the Gaussian-elimination-algorithm on it. The columns which you will obtain then form a basis of the image.
Need help
So the basis of the Im(T) is the vector columns of the matrix.
 [/INST] You can try this way: choose basis $\displaystyle \{1,x,x^2,x^3\}$ of $P_3$. Then check:
$$T[x^3]=1$$
$$T[x^2]=x$$
$$T[x]=x^2$$
$$T[1]=x^3$$
So first element of basis is transofmed to fourth  by $T$, second to third, third to second, fourth to first. So the matrix representation is:
$$T=\begin{bmatrix}0 && 0 && 0 && 1 \\ 0 && 0 && 1 && 0 \\ 0 && 1 && 0 && 0 \\ 1 && 0 && 0 && 0\end{bmatrix}$$.
Now you can check that $\det T \neq 0$, so Ker(T)=$\{0\}$, so basis of Ker(T) is $\{0\}$, Im(T)=$P_3$, so basis Im(T) is $P_3$, for example, $\displaystyle \{1,x,x^2,x^3\}$.
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