Let V be the function space of all polynomials up to degree 2. Then define the following two linear transformations on $V$:
For every $f(x)\in V$:
$S(f(x))=x^2\cdot f(\frac{-1}{x})$ and $T(f(x))=f(x+a)$, where $a$ denotes some complex number.
I further define the following transformation:
$U(f(x))=S(T(f(x)))$.
I now want to find the characteristic polynomial of $U$, and I don't understand the question. Because I thought that a characteristic polynomial always just exists when we are dealing with matrices.
Furthermore, I want the find the $a$ such that $U$ becomes non-diagonisable, which I also don't understand.
Maybe it is trivial.... But I've never dealt with this before.
Hope someone can help me.
Edit: thank. This is 4 years ago for me. I have been reading up on it now. I found the two matrices to be:
S=$\begin{bmatrix}0&0&1\\0&-1&0\\1&0&0 \end{bmatrix}$ and T=$\begin{bmatrix}1&a&a^2\\0&1&2a\\0&0&1 \end{bmatrix}$
Then for the transformation $U=S\circ T$, I find that the matrix of basis vector of V for $U$ is given as
$U=S\cdot T = \begin{bmatrix}0&0&1\\0&-1&-2a\\1&a&a^2 \end{bmatrix}$.
If I want to diagonalize $U$ as well, how can I do that without knowing the an invertable matrix used for this diagonalisation. Further, what characterises a matrix as undiagonalisable? I it if the $Det(U)=0$?
Hint
$V$ is a vector space over the real. A basis is $B = (1,x,x^2)$. You have to find the matrix of the linear maps $S,T$ in $B$.
For example $x^2 = 0 \cdot 1 + 0 \cdot x +1 \cdot x^2$ and $$T(x^2) = (x+a)^2= a^2 \cdot 1 + 2a \cdot x + 1 \cdot x^2.$$ Therefore the last column of the matrix of T in the basis $B$ is $$\begin{pmatrix}a^2 \\ 2a\\ 1\end{pmatrix}$$
Look carefully to understand why if this is not immediate to you... you'll have learn something interesting!
Now your turn to find the other columns of $T$ and the matrix $S$!
Based on that finding the matrix of $U$ is just having fun (!?!) multiplying matrices...
Have fun!