# the find the trace of $A^{102}$?

if $$A = \begin{bmatrix} 1 &0 &0 \\i& \frac{-1+ i\sqrt 3}{2} &0\\0&1+2i &\frac{-1- i\sqrt 3}{2} \end{bmatrix}$$.Then the find the trace of $$A^{102}$$?

My attempt : i know that eigenvalue of A are $$1,w$$ and $$w^2$$

As im not able to proceed further pliz help me

• JCF is way too much for this. Evaluate $\;A^2,\,A^3\;$ ...and try to make an educated guess, which you can prove formally by induction. – DonAntonio Nov 18 '18 at 13:02

## 5 Answers

The matrix $$A$$ has eigenvalues the cube roots of unity $$1, j=\mathrm e^{\tfrac{2i\pi}3}$$ and $$\bar{\mkern-1muj\mkern2mu}$$, and its characteristic polynomial is $$X^3-1$$. So, by Hamilton-Cayley, we have $$\;A^3=I$$, therefore $$\;A^{102}=(A^3)^{34}=I$$ and $$\operatorname{Tr}\bigl(A^{102}\bigr)=\operatorname{Tr}(I)=3.$$

It's not necessary to use any theorems about eigenvalues here. It's easy to see (and straightforward to prove, by induction) that powers of a lower triangular matrix are all lower triangular, and that the diagonal elements of $$A^n$$ are the $$n$$th powers of the diagonal powers of $$A$$. Since $$\left(-1\pm i\sqrt3\over2\right)^3=1$$, and since $$102=3\cdot34$$, it follows that the diagonal elements of $$A^{102}$$ are all $$1$$, so the trace of $$A^{102}$$ is $$1+1+1=3$$.

D=$$\begin{pmatrix}1 & 0 & 0\\0 & \omega & 0\\0 & 0 & \omega^2\end{pmatrix}$$.

We know that A is similar to D. So there exist an invertible matrix P, such that $$A=PDP^{-1}$$.

Now $$A^2=PDP^{-1}*(PDP^{-1})=PD^2P^{-1}$$

Similarly, $$A^k=PD^kP^{-1}$$

$$A^{102}=PD^{102}P^{-1}$$

So $$A^{102}$$ and $$D^{102}$$ are similar and there trace are equal.

So $$trace(A^{102})=trace(D^{102})=1^{102}+\omega^{102}+\omega^{204}=1+1+1=3$$

In general, for any polynomial p(X) and any square matrix A, the eigenvalues of p(A) are p($$\lambda$$), where $$\lambda$$ are the eigenvalues of A.

In your case, the polynomial is $$X^{102}$$ and therefore the eigenvalues of $$A^{102}$$ will be $$1, w^{102} and (w^2)^{102}=w^{204}$$ , since as you noted, the eigenvalues of A can be read off the main diagonal and are $$1,w,w^2$$

Since w is a third root of the unity, $$w^3=1$$ and so $$w^{102}=w^{204}=1$$ and therefore the trace is the sum of the three values namely $$1+1+1=3$$

By the Jordan decomposition theorem there exists a matrix $$B \in \mathbb{C}^{3 \times 3}$$, so that the matrix $$B^{-1} A B$$ has Jordan normal form, $$J(A)$$.

From there follows $$A = B^{-1} J(A) B$$ and so $$$$A^{102} = \left(B^{-1} J(A) B \right)\left(B^{-1} J(A) B\right) \ldots \left(B^{-1} J(A) B\right)$$$$ Notice, that the adjacent $$B$$ and $$B^{-1}$$ always cancel, so we obtain $$A^{102} = B^{-1} J(A)^{102} B$$.

You have already found the eigenvectors, so finding the Jordan normal form isn't far away.