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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
You know $A = PDP^{-1}$ for some diagonal $D$ and invertible $P$. The columns of $P$ are eigenvectors corresponding to the eigenvalues on the diagonal of $D$. The range of $L$ is therefore the span of the columns of $P$ corresponding to nonzero eigenvalues. (Note that the size of this basis is $k$, the number of nonzero eigenvalues.) This is an answer to your question, although it is contingent on you having $A = PDP^{-1}$ already.
A more concrete procedure is the following. For each nonzero eigenvalue $\lambda$, suppose it has multiplicity $m_\lambda$. Find a basis for the nullspace of $A - \lambda I$ (which has dimension $m_\lambda$), which produces a linearly independent set of $\lambda$-eigenvectors. Doing this for each nonzero $\lambda$ and collecting the various bases yields the desired basis for the range of $L$.
