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Let $ \alpha = e^{\frac{2\pi \iota}{5}}$ and the matrix $ M= \begin{pmatrix}1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\ 0 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\ 0 & 0 & \alpha^2 & \alpha^3 & \alpha^4 \\ 0 & 0 & 0 & \alpha^3 & \alpha^4\\ 0 & 0 & 0 & 0 & \alpha^4 \end{pmatrix}$

Then the trace of the matrix $I + M + M^2$ is

  1. $-5$;
  2. $0$;
  3. $3$;
  4. $5$.

I am stuck on this problem. Can anyone help me please?

I got trace of the matrix $\operatorname{tr}(I+M+M^2) = 7 + \alpha + 2 \alpha^2 + \alpha^3 + 2 \alpha^4 + \alpha^6 +\alpha^8.$ Now what to do?

  • 0
    $tr(I+M+M^2) = 7 + \alpha + 2 \alpha^2 + \alpha^3 + 2 \alpha^4 + \alpha^6 +\alpha^8 $ now using $\alpha^5 =1,1+\alpha+\alpha^2+\alpha^3+\alpha^4=0 $ its 5.2012-12-31

1 Answers 1

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Note that the trace of $M$ is $0$, since $1+\alpha+\alpha^2+\alpha^3+\alpha^4= 0$.

Also $M$ is upper triangular so that $M^2$ has diagonal elements which are just the square of the diagonal elements of $M$, i.e. $1,\alpha^2, \alpha^4, \alpha^6, \alpha^8$.

Using the fact that $\alpha^5 = 1$ we see that the trace of $M^2$ is again $0$.

Thus tr$(I+M+M^2)$ = tr$(I)$ = 5.