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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

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  • $\begingroup$ 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. $\endgroup$ – DonAntonio Nov 18 '18 at 13:02
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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 $$\operatorname{Tr}\bigl(A^{102}\bigr)=\operatorname{Tr}(I)=3.$$

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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 $1+1+1=3$.

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

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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 $1+1+1=3$

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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 \begin{equation} A^{102} = \left(B^{-1} J(A) B \right)\left(B^{-1} J(A) B\right) \ldots \left(B^{-1} J(A) B\right) \end{equation} 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.

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