Finding the trace of $T^4$ and $T^2$? 
Q. Let $V$ denote the (complex) vector space of complex polynomials of degree at most $9$ and consider the linear operator $T:V \to V$ defined by $$T(a_0+a_1 x+a_2 x^2+\cdots+a_9 x^9)=a_0 +(a_2 x +a_1 x^2)+(a_4 x^3+ a_5 x^4 + a_3 x^5)+(a_7 x^6 + a_8 x^7 + a_9 x^8 + a_6 x^9).$$ 
(a) What is the trace of $T^4$?
(b) What is the trace of $T^2$?

My approach :
First off we form matrix of $T$ under the standard basis $\{1,x,x^2,...,x^9\}.$ We get,
$$
    \left(\begin{matrix}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0  \\
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}\right)
$$
Now $\text {tr}(T)=1$. We calculate $\det T$ by observing number of row exchanges that convert it to an identity matrix. Number of such exchanges is $6$. Thus $\det T=(-1)^6=1.$
We have relation of eigenvalues with trace and determinants as follows,
$\lambda_1+\lambda_2+...+\lambda_{10}=\text {tr} (T)=1$ and $\lambda_1 \lambda_2 \cdots \lambda_{10}=\det T=1.$
Also eigenvalues of $T^4$ and $T^2$ are ${\lambda_1}^4,{\lambda_2}^4,...,{\lambda_{10}}^4$ and ${\lambda_1}^2,{\lambda_2}^2,...,{\lambda_{10}}^2$ respectively.
In the $3 \times 3$ and $4 \times 4$ cases of this problem, I found $1$ as a common eigenvalue and other eigenvalues to be either $-1$ or $\pm i$ or $\frac {-1 \pm \sqrt 3i}2.$ So eigenvalues are coming out to be $n-$th roots of unity but I am not able to find a pattern among them.
My plan is using $\text {tr} (T^2)={\lambda_1}^2 + {\lambda_2}^2 +...+{\lambda_{10}}^2$ and $\text {tr}(T^4)={\lambda_1}^4+{\lambda_2}^4+...+{\lambda_{10}}^4.$
Knowing all these things is not helping me to find a way to tackle the problem. Is there some concept that I am missing to apply?
Moreover are there any shorter elegant ways to approach this problem?
 A: Notice that $T$ just acts as $x^i \mapsto x^{\sigma(i)}$, where $\sigma$ is a permutation on $\{0, \ldots, 9\}$ given as:
$$\sigma = (1\,2)(3\,5\,4)(6\,9\,8\,7)$$
So $T^2$ acts as $x^i \mapsto x^{\sigma^2(i)}$, where $$\sigma^2 = \sigma \circ \sigma = (1\,2)^2(3\,5\,4)^2(6\,9\,8\,7)^2 = (3\,5\,4)^2(6\,9\,8\,7)^2$$
so
$$T^2(a_0+a_1 x+a_2 x^2+\cdots+a_9 x^9)=a_0 +a_1 x +a_2 x^2+(a_5 x^3+ a_3 x^4 + a_4 x^5)+ (a_8 x^6 + a_9 x^7 + a_6x^8 + a_7 x^9)$$
The trace in this case is just the number of elements $x_i$ such that $x^i \mapsto x^{i}$, that is the fixed points of our permutation $\sigma^2$. So, $\operatorname{Tr} T^2 = 3$, since the fixed points of $\sigma^2$ are $0$, $1$, $2$.
Similarly $T^4$ acts as $x^i \mapsto x^{\sigma^4(i)}$, where $$\sigma^4 = (1\,2)^4(3\,5\,4)^4(6\,9\,8\,7)^4 = (3\,5\,4)$$
so
$$T^4(a_0+a_1 x+a_2 x^2+\cdots+a_9 x^9)=a_0 +a_1 x +a_2 x^2+(a_4 x^3+ a_5 x^4 + a_3 x^5)+a_6 x^6 + a_7 x^7 + a_8 x^8 + a_9 x^9$$
We read $\operatorname{Tr} T^4 = 7$, since the fixed points of $\sigma^4$ are $0, 1, 2, 6, 7, 8, 9$.
A: Hint: The matrix is diagonal by block
$$ T = 
\begin{pmatrix} 
A & 0 & 0 & 0 \\
0 & B & 0 & 0 \\
0 & 0 & C & 0 \\
0 & 0 & 0 & D \\
\end{pmatrix}
$$
with $$ A = I_{1,1}, B = 
\begin{pmatrix} 
0 & 1  \\
1 & 0  \\
\end{pmatrix}, C = \begin{pmatrix} 
0 & 1 & 0 \\
0 & 0 & 1\\
1 & 0 & 0 \\
\end{pmatrix}, D = \begin{pmatrix} 
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
\end{pmatrix}$$
since $T$ is diagonal by block :
$$T^n = \begin{pmatrix} 
A^n & 0 & 0 & 0 \\
0 & B^n & 0 & 0 \\
0 & 0 & C^n & 0 \\
0 & 0 & 0 & D^n \\
\end{pmatrix}$$
$$Tr(T^n)=Tr(A^n)+Tr(B^n)+Tr(C^n)+Tr(D^n)$$
$$Tr(T^2) = 1+2+0+0 = 3$$
$$Tr(T^4) = 1+2+0+4 = 7$$
You can think of this in term of application and subspaces : there are 4 subspaces of $V$ that are stable to $T$, if you define restriction of $T$ in these subspaces you have much simpler applications (that are only circle permutation moreover)
