Let \(V\) be an \(n\)-dimensional vector space, and let \(T\) be a linear operator on \(V\). Suppose that \(T\) is diagonalizable.
If \(T\) has a cyclic vector, show that \(T\) has \(n\) distinct characteristic values.
If \(T\) has \(n\) distinct characteristic values, and if \(\{\alpha_1, \ldots, \alpha_n\}\) is a basis of characteristic vectors for \(T\), show that \(\alpha = \alpha_1 + \cdots + \alpha_n\) is a cyclic vector for \(T\).
Solution
1. A diagonalizable operator with a cyclic vector must have \(n\) distinct characteristic values.
Suppose \(T\) has a cyclic vector \(\alpha\). Then \[ \alpha,T\alpha,\ldots,T^{n-1}\alpha \] span the \(n\)-dimensional space \(V\) and hence form a basis. Therefore no nonzero polynomial of degree less than \(n\) can annihilate \(\alpha\); such a relation would be a nontrivial linear dependence among these basis vectors. On the other hand, \(T^n\alpha\) is a linear combination of the displayed basis, so some monic polynomial of degree \(n\) annihilates \(\alpha\). Hence the \(T\)-annihilator of \(\alpha\) has degree \(n\).
For a cyclic vector its annihilator equals the minimal polynomial of \(T\) [@HK71], so \[ \deg m_T=n. \] If \(T\) is diagonalizable and its distinct characteristic values are \(\lambda_1,\ldots,\lambda_r\), then \[ m_T(x)=\prod_{i=1}^r(x-\lambda_i). \] Thus \(r=\deg m_T=n\). Therefore \(T\) has \(n\) distinct characteristic values.
2. The sum of an eigenbasis is cyclic when the eigenvalues are distinct.
Let \[ T\alpha_i=\lambda_i\alpha_i, \qquad i=1,\ldots,n, \] where the \(\lambda_i\) are distinct, and put \[ \alpha=\alpha_1+\cdots+\alpha_n. \] Then for \(0\le k\le n-1\), \[ T^k\alpha=\sum_{i=1}^n\lambda_i^k\alpha_i. \] Relative to the eigenbasis \((\alpha_1,\ldots,\alpha_n)\), the coordinate columns of \[ \alpha,T\alpha,\ldots,T^{n-1}\alpha \] form the Vandermonde matrix \[ \begin{pmatrix} 1&\lambda_1&\lambda_1^2&\cdots&\lambda_1^{n-1}\\ 1&\lambda_2&\lambda_2^2&\cdots&\lambda_2^{n-1}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\lambda_n&\lambda_n^2&\cdots&\lambda_n^{n-1} \end{pmatrix}. \] Its determinant is \[ \prod_{1\le i<j\le n}(\lambda_j-\lambda_i)\ne0 \] because the characteristic values are distinct. Hence those \(n\) vectors form a basis of \(V\). Therefore \(\alpha\) is a cyclic vector for \(T\).