Here is a question I encountered in my homework.
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 n distinct characteristic values, and if {$\alpha_{1}$, ..., $\alpha_{n}$} is a basis of characteristic vectors for T, show that $\alpha$ = $\alpha_{1}$ + ... + $\alpha_{n}$ is a cyclic vector for T.
I want to ask what's the meaning of "{$\alpha_{1}$, ..., $\alpha_{n}$} is a basis of characteristic vectors for T" ? Does it means {$\alpha_{1}$, ..., $\alpha_{n}$} are characteristic vectors?
We say that $\alpha$ is a characteristic vector (AKA eigenvector or proper vector) if there is a scalar (i.e. a number as opposed to a vector) $\lambda$ for which $T(\alpha) = \lambda \alpha$. The number $\lambda$ is called the "characteristic value" associated with the characteristic vector $\alpha$.
The phrase "$\{\alpha_{1},\dots , \alpha_{n}\}$ is a basis of characteristic vectors for $T$" means that $\{\alpha_{1},\dots , \alpha_{n}\}$ is a basis and that each vector $\alpha_i$ is a characteristic vector.
This is fairly standard English usage. Another example of such a phrase is "the object on the table is a wheel of cheese," which means that the object on the table is a wheel, and the material that comprises the wheel is cheese.