Suppose $AB=BA$ and $x$ is an eigenvector of $A$ with geometric multiplicity $k$, can $Span\{x, Bx, ...\}$ be represented?

Question

Suppose $A$ and $B$ are $n \times n$ commuting matrix: $AB=BA$. $x$ is an eigenvector of $A$ with the eigenvalue $\lambda$ of geometric multiplicity $k$ i.e. $\ker(\lambda I-A)$ is of dimension $k$. Clearly each nonzero vector $B^i x$ is the eigenvector of $A$. I want to study of the representation of the space $$Span\{x,Bx,B^2x,\ldots\}=\{\Sigma a_iB^{i-1}x \quad \text{where the sum is of finite terms.}\}$$ by finite terms of $\{B^i x\}$. To be more precisely, is $\{x,Bx,...,B^{k-1}x\}$ the basis of this vector space ( or $\{x,Bx,...,B^m x\}$ is a basis for another integer $m$)? I fail to see both the linear independence and the representability of this group of vectors in the space.

In the case where both $A$ and $B$ have $n$ different eigenvalues (or can be diagonalized), $A$ and $B$ can be diagonalized at the same time and the statement above is true. But in general case, it is not true that $A$ and $B$ can be normalized to the Jordan Normal Forms at the same time. That is, $$\exists P \in GL_n \quad s.t. P^{-1}AP=J_A \quad P^{-1}BP=J_B$$ can be false.

Hence I feel difficult to answer this question though Jordan Normal Form.

Any help will be highly appreciated.

Answer 1

Let $m$ denote the minimal index such that $$B^{m+1}x \in \operatorname{span}_F\{x, Bx,\ldots, B^mx\}.$$

Then as a result, $$i \leq m \implies B^ix \not\in \operatorname{span}_F\{x,Bx, \ldots, B^{i-1}x\},$$ and in particular, the set $\mathcal{B} = \{x, Bx, \ldots, B^mx\}$ is linearly independent over $F$.

Since $\mathcal{B}$ is a subset of $\ker(\lambda I - A)$, which has dimension $k$, we conclude $m+1 = |\mathcal{B}| \leq k$, so $m < k$.

If $m = k-1$, then $\mathcal{B}$ spans $\ker(\lambda I - A)$, and therefore is a basis of that subspace. Otherwise, $\operatorname{span}_F(\mathcal{B})$ is some proper subspace of $\ker(\lambda I - A)$, and $\mathcal{B}$ is a basis of that subspace.

Answer 2

I think the answer offered by Brian Moehring is right. But a lemma that does not seem so trivial is needed.

**lemma:**If the integer $m$ is minimal such that $\{x, Bx, ... ,B^m x\}$ is linearly dependent, then for all $l \ge m$, $B^l x$ can be represented by $\{x, Bx,..., B^{m-1}x\}$.

Proof: One can use induction to decrease the degree of $B^l x$ to $B^{l-1}x$:

If$$B^{m}x = \Sigma_{i=0}^{m-1}a_iB^{i}x$$ then$$B^{l}x = B^{m+(l-m)x}=\Sigma a_{i=0}^{m-1}a_iB^{i+l-m}$$ whose highest degree is of $l-1$.

Note that $m$ is not necessary to be the geometric multiplicity of the eigenvector space of $A$ with respect to $\lambda$. For example, let $A=I$ then for all square matrix $B$, $AB=BA$ and $x$ can be any vector, while $\dim \square$ is not necessary to be $k$.