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

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 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 $1+1+1=3$

1 Votes

Sorin Tirc
November 18, 2018 13:01 PM

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.

1 Votes

Viktor Glombik
November 18, 2018 13:04 PM

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

0 Votes

Jimmy
November 18, 2018 13:12 PM