$V$ is direct sum of $1$ dimensional subspaces invariant with respect to $A$ and $B$ simultaneously where $A^{m}=B^{n}=I_{V}$ and $AB=BA$.

Question

Let $A,B$ are commuting linear operator on a finite dimensional vector space $V$ over the field $\mathbb{C}$ such that $A^{m}=B^{n}=I_{V}$ for some positive integers $m,n$. Prove that $V$ is direct sum of $1$ dimensional subspaces invariant with respect to $A$ and $B$ simultaneously.

Since $\mathbb{C}$ is algebraically closed, polynomials of the form $X^{r}-1$ with $r\in \mathbb{N}$, splits over $\mathbb{C}$ and, moreover, separable polynomials. Thus both $A$ and $B$ are diagonalizable and $V$ becomes direct sum of eigenspaces of both $A$ and $B$. Also $AB=BA$ $\implies A(B-\lambda I)=(B-\lambda I)A$ for all $\lambda \in \mathbb{C}$, in particular this holds when $\lambda$ is an eigenvalue of $B$. This shows that eigenspace of $B$ is invariant under $A$, and similar thing holds for $B$. But how do I split further the eiegenspaces in one dimensional invariant spaces? which looks unlikely. Is there any other way we can proceed? Thanks.

Answer 1

As you have noted, the eigenspaces of $B$ are invariant under $A$ (and vice versa). Now, consider any eigenspace $\ker(B - \lambda I)$ (of $B$ associated with eigenvalue $\lambda$). Because $A$ is diagonalizable, so is $A|_{\ker(B - \lambda I)}$, which means that $\ker(B - \lambda I)$ can be decomposed into a sum of eigenspaces of $A$. Note that any subspace of an eigenspace is invariant, so these subspaces are invariant with respect to both $A$ and $B$. By doing this for all eigenspaces of $B$, we can decompose $V$ into a sum of spaces that are simultaneously invariant under both $A$ and $B$.

Because all of these subspaces are contained in subspaces of $A$ and $B$, they can be decomposed into one-dimensional eigenspaces, and it is guaranteed that the resulting 1-dimensional subspaces are invariant under both $A$ and $B$.

Answer 2

You are at the stage where you know that both $A,B$ are diagonalizable, and what you want to show is that they are simultaneously diagonalizable.

Because $A$ is diagonalizable, we can write $$ V=\bigoplus_1^rV_r $$ where each $V_r$ is an eigenspace for $A$ with eigenvalue $\lambda_r$. Because $B$ commutes with $A$, $BV_r\subset V_r$. Over a subspace, the minimal polynomial of $B$ cannot acquire irreducible factors (because it has to divide the minimal polynomial of $B$ over $V$). So $B$ is diagonalizable over $V_r$, and then we can write $V_r=\bigoplus_j V_{r,j}$ where $V_{r,j}\subset V_r$ is a one-dimensional eigenspace for $B$. Then $$ V=\bigoplus_{j,r}V_{r,j} $$ is a decomposition of $V$ into a direct sum of common invariant subspaces for $A$ and $B$.