2
$\begingroup$

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.

  • 0
    You're welcome!2012-11-25

1 Answers 1

2

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.