Let \(A\) be an \(n\times n\) matrix over a field \(F\) such that \(A\) is diagonalizable. Prove that the following are equivalent:
There is a vector \(v\in F^n\) such that \(v, Av, \cdots A^{n-1}v\) is a basis for \(F^n\).
The eigenvalues of \(A\) are distinct.
Solution
1. Suppose first that \(A\) has \(n\) distinct eigenvalues \(\lambda_1,\dots,\lambda_n\) with eigenbasis \(e_1,\dots,e_n\). Set\[ v=e_1+\cdots+e_n. \]Then \(v,Av,\dots,A^{n-1}v\) is a basis.
Relative to the eigenbasis, the columns of \[ [v\ Av\ \cdots\ A^{n-1}v] \] form the Vandermonde matrix \[ \begin{pmatrix} 1&\lambda_1&\cdots&\lambda_1^{n-1}\\ \vdots&\vdots&&\vdots\\ 1&\lambda_n&\cdots&\lambda_n^{n-1} \end{pmatrix}. \] Its determinant is \[ \prod_{i<j}(\lambda_j-\lambda_i), \] which is nonzero because the eigenvalues are distinct. Hence the displayed vectors are linearly independent and therefore form a basis.
2. Conversely, suppose there is \(v\) such that\[ v,Av,\dots,A^{n-1}v \]is a basis. Then the minimal polynomial of \(A\) has degree \(n\).
If a nonzero polynomial \[ p(t)=c_0+c_1t+\cdots+c_dt^d \] of degree \(d<n\) satisfied \(p(A)=0\), then \[ c_0v+c_1Av+\cdots+c_dA^dv=0, \] contradicting linear independence of the basis vectors. Thus no annihilating polynomial of degree less than \(n\) exists. Since the characteristic polynomial has degree \(n\) and annihilates \(A\), the minimal polynomial has degree exactly \(n\).
3. Since \(A\) is diagonalizable, its minimal polynomial is a product of distinct linear factors, one for each distinct eigenvalue.
For a diagonalizable operator, the minimal polynomial is \[ \prod_{\lambda\in\operatorname{Spec}(A)}(t-\lambda). \] Therefore its degree equals the number of distinct eigenvalues. By 2 this degree is \(n\), so \(A\) has \(n\) distinct eigenvalues.
Thus the two conditions are equivalent.