# Problems related with a linear transformation

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. – levap Nov 23 '12 at 18:16
This happens to be an example of en.wikipedia.org/wiki/Companion_matrix and so the characteristic polynomial is particularly easy to calculate. – Cocopuffs Nov 23 '12 at 18:21
@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? – learner Nov 23 '12 at 19:47
You are right. But did you try also the $(v_1, v_2)$ case? It might be different! – levap Nov 24 '12 at 19:55
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. – levap Nov 25 '12 at 13:25

The comment about companion matrix lets you write down the charactistic polynomial easily, and even if you know nothing about companion matrices you should be able to find the minimal polynomial of $T$ easily (that polynomial evaluated in $T$ sould send $v_1$ to $0$), which happens to be the same as the charactistic polynomial (well, it is not such a coincidence given Cayley-Hamilton). You may recognise that polynomial as a particular one, and conclude from there.
There's an easier way though, a bit ad hoc. Just compute $T^2(v_{16})$, and from there find a non-minimal polynomial that annihilates $T$, and whose rational roots are easy to find. None of these roots happens to be an eigenvalue of $T$, which should lead you to making the right choice.