# 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?

• Surely the fact that $T$ is a permutation matrix, and the cycle decomposition of the associated permutation, must help. Nov 23 '17 at 13:18
• @kimchilover I didn't understand. How to use cycle decomposition here? How will it reflect in the trace? Nov 23 '17 at 13:26
• "It will reflect the trace", because we then know how the powers $T^k$ look like - and reading off the trace will be trivial. Nov 23 '17 at 13:31
• The cycle type of the original permutation $T$ is$1^1 2^1 3^1 4^1$. The type of $T^2$ is thus $1^3 2^2 3^1$, and of $T^4$ is 1+1+1 $1^7 3^1$. The trace of a permutation is the number of fixed points, that is, of $1$-cycles. Nov 23 '17 at 13:37
• @kimchilover Ah! Got your idea. It's so helpful. Couldn't imagine this way myself. Nov 23 '17 at 15:41

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 $0, 1, 2, 6, 7, 8, 9$.

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)

• Can you expand this hint a bit? Surely we can write down the JCF of $T$. But won't we need to calculate eigenvalues first? Nov 23 '17 at 13:31
• @Mathmore I don't know what JCF means but I expanded my answer Nov 23 '17 at 13:44
• Oh I am sorry. I mean't Jordan Canonical Form which is also called as Jordan Normal Form. Nov 23 '17 at 13:45