suppose $A$ is an $n \times n$ diagonalizable matrix with exactly $k$ non-zero eigenvalues, and let $L(\vec{x})=A\vec{x}$. How would you find a basis {$\vec{v_1},...,\vec{v_k}$} for $Range(L)$ such that $\vec{v_1},...,\vec{v_k}$ are all eigenvecgtors of A?
I know that since $A$ is diagonalizable there exists a basis {$\vec{v_1},...,\vec{v_n}$} for $\Bbb{R^n}$ of eigenvectors and also that $A\vec{x}=P^{-1}DP\vec{x}$ for some invertible matrix $P$ and the diagonal matrix D