# $A$ is an $n \times n$ diagonalizable matrix with exactly $k$ non-zero eigenvalues, find a basis for $Range(L)$

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

• When you say $k$ non-zero eigenvalues, do you mean counted with multiplicity? Apr 6, 2018 at 20:59
• just that there are k eigenvalues with all their multiplicites adding to n Apr 6, 2018 at 21:00
• In that case, the dimension of the range of $L$ could be greater than $k$. For instance, if $A$ is the identity matrix. Apr 6, 2018 at 21:01

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 $L$.
• @Skrrrrrtttt The eigenvectors themselves form a basis for $\mathbb{R}^n$, so their images under $A$ will span the image. Apr 6, 2018 at 22:15