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I have been trying to solve the following problem.

Let $\{v_{1},v_{2},....,v_{16}\}$ be an ordered basis for $V=\mathbb{C}^{16}$.If $T$ is a linear transformation on $V$ defined by $T(v_{i})=v_{i+1}$ for $1\leq i\leq 15$ and $T(v_{16})=-(v_{1}+v_{2}+....+v_{16}).$

Then which of the following is/are true?

(a)$T$ is singular with rational eigenvalues,

(b)$T$ is singular but has no rational eigenvalues,

(c)$T$ is regular(invertible) with rational eigenvalues,

(d)$T$ is regular but has no rational eigenvalues.

Could someone point me in the right direction(e.g. a certain theorem or property I have to use?) Any kind of hints will be helpful.

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    You can try first the analogous $2 \times 2$ or $3 \times 3$ case. That is, work with $(v_1, v_2)$ and $(v_1, v_2, v_3)$ and see what you can make out of it.2012-11-23
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    This happens to be an example of http://en.wikipedia.org/wiki/Companion_matrix and so the characteristic polynomial is particularly easy to calculate.2012-11-23
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    @levap I have tried the analogous 3 X 3 case.I worked with (v_{1},v_{2},v_{3}) and see that matrix thus formed is invertible and i have also got rational eigen values. Am i right,sir?2012-11-23
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    You are right. But did you try also the $(v_1, v_2)$ case? It might be different!2012-11-24
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    @levap In case of analogous 2 X 2 case [that is for (v_{1},v_{2})], we see that the matrix thus formed is invertible but in this case, we do not have any rational eigen values. Is not it,sir?2012-11-25
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    You are correct! So, can you generalize it to $16 \times 16$? Will it be like the $2 \times 2$ case or like the $3 \times 3$ case?2012-11-25
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    @levap In fact,i have also tried 4 X 4 case.I noticed that the matrix thus formed is invertible without having rational eigen values. Thus we see if the order of the matrix is even then we get invertible matrix without having rational eigen values.So, in this case,the order of the matrix(I mean the matrix representation of T) is even. So,i think option (4) will be correct. Is not it,sir?2012-11-25
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    Bingo. To give some details, the characteristic polynomial of the $n \times n$ case will be $t^n + t^{n-1} + ... + t + 1$. What are the roots of this polynomial? As $(t - 1)(t^n + t^{n-1} + ... + 1) = t^{n+1} - 1$, the roots are the $n+1$ roots of unity, except $1$. If $n$ is odd, then $n + 1$ is even and so $-1$ is a rational root. If $n$ is even, then $n + 1$ is odd and all the roots $e^{\frac{l 2 \pi i}{n + 1}}$ are complex numbers with non-zero imaginary part and so there are no rational eigenvalues.2012-11-25
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    @levap thank you so much sir for clarifying it and for your time.2012-11-25
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    You're welcome!2012-11-25

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