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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?

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1  
If $A$ and $B$ are matrices, what do you know about $tr(AB)$? –  Calvin Lin Dec 31 '12 at 6:49
2  
It will be good if you give what you tried and where exactly you got struck. –  Ram Dec 31 '12 at 6:52
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@Prasanta Can you explain what happens to the elements on the diagonal when you multiply 2 upper triangular matrices? For example, can you tell me what are the elements on the diagonal of $M^2$ without actually doing the multiplication? –  Calvin Lin Dec 31 '12 at 6:59
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@Prasanta Use that fact that since $\alpha = 2^{ \frac {2\pi i}{5}}$ is a fifth root of unity, so $\alpha^5 = 1, 1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 = 0$ –  Calvin Lin Dec 31 '12 at 7:01
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@Prasanta Why don't you write up your solution and post it, so others can read it? –  Calvin Lin Dec 31 '12 at 7:06
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1 Answer

up vote 7 down vote accepted

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.

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