How to find if $T: P_{3}(\mathbb{R}) \Rightarrow P_{3}(\mathbb{R}) $ is diagonalizable? 
How to find if $T: P_{3}(\mathbb{R}) \to P_{3}(\mathbb{R})$ is diagonalizable, and if so how to find a basis containing only eigenvectors of T?

Let T be defined by 
$T(ax^{3}+bx^{3}+cx+d) = dx^{3}+cx^{3}+bx+a$
I am having trouble with this, it seems pretty trivial, I don't know what I am doing wrong.
I started by decomposing it
$T(ax^{3}+bx^{3}+cx+d) = \\dx^{3}+cx^{3}+bx+a =\\(a/a)dx^{3}+(b/b)cx^{3}+(c/c)bx+(d/d)a =\\
(d/a)ax^{3}+(c/b)bx^{3}+(b/c)cx+(a/d)d\\$
Thus 
$T(ax^{3}+bx^{3}+cx+d) = 
(d/a)ax^{3}+(c/b)bx^{3}+(b/c)cx+(a/d)d$
Which yields a matrix of transformation:
$ \left( \begin{array}{ccc}
d/a & 0 & 0 & 0 \\
0 & c/b & 0 & 0 \\
0 & 0 & b/c & 0 \\
0 & 0 & 0 & a/d \end{array} \right) 
$
So far so good, but now... I believe I am doing something wrong because no matter what I try I can't get a basis. The eigenvalues are clearly all the items on the diagonal...
I am confused, how do I find a basis for this?
 A: I'd say you set up your matrix incorrectly, because the fact that you're dividing by the coefficients means you can't evaluate $T(x^3)$ using your matrix. ($M_{1,1}=1/0$.) Also, your matrix is not "universal": If you multiply $\pmatrix{d/a&0&0&0\\ 0&c/b&0&0\\ 0&0&b/c&0\\ 0&0&0&a/d} \cdot \pmatrix{3\\ 2\\1\\0}$, you get $\pmatrix{3d/a\\ 2c/b\\ b/c\\ 0}\not=\pmatrix{0\\1\\2\\3}$.
The matrix should map coordinates of a polynomial $p$ to coordinates of a polynomial $p$ when you're dealing with vector spaces other than $R^n$.
If you are using the standard ordered basis $B=(x^3,x^2,x,1)$, then the polynomial $p(x)=ax^3+bx^2+cx+d$ is represented by the vector $\pmatrix{a\\ b\\ c\\ d}$. (This is also the coordinate vector of $p$ with respect to $B$.) The matrix representing $T$ is then $\left(T(x^3)_B~~T(x^2)_B~~T(x)_B~~T(1)_B\vphantom{\Bigg|}\right)$. (The subscript $_B$ means "coordinates with respect to $B$.) Then $T(x^3)_B = (1)_B = \pmatrix{0\\ 0\\ 0\\ 1}$, according to your definition, and your matrix becomes $\pmatrix{0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0}$. Now the question is whether this matrix is diagonalizable.
