Let \(T\) be a linear operator on \(F^2\). Prove that any non-zero vector which is not a characteristic vector for \(T\) is a cyclic vector for \(T\). Hence, prove that either \(T\) has a cyclic vector or \(T\) is a scalar multiple of the identity operator.
Solution
1. Every nonzero non-characteristic vector is cyclic.
Let \(0\ne\alpha\in F^2\) and suppose \(\alpha\) is not a characteristic vector for \(T\). Then \(T\alpha\) is not a scalar multiple of \(\alpha\), so \[ \alpha,T\alpha \] are linearly independent. Since \(F^2\) has dimension two, they form a basis. Therefore the \(T\)-cyclic subspace generated by \(\alpha\), which contains \(\alpha\) and \(T\alpha\), is all of \(F^2\). Hence \(\alpha\) is cyclic [@HK71].
2. If there is no cyclic vector, then \(T\) is scalar.
Assume \(T\) has no cyclic vector. By step 1, every nonzero vector is then a characteristic vector. Choose a basis \(u,v\). There are scalars \(a,b,c\) with \[ Tu=au, \qquad Tv=bv, \qquad T(u+v)=c(u+v). \] By linearity, \[ au+bv=cu+cv. \] Independence of \(u,v\) gives \(a=b=c\). Thus \(T\) has the same scalar action on a basis and hence \[ T=aI. \] Consequently either \(T\) has a cyclic vector or it is a scalar multiple of the identity.