Suppose $\mathcal{A}$ is a linear operator on some vector space $V$, and suppose $U$ is a $\mathcal{A}$-invariant subspace of $V$. Does there necessarily exist a corresponding linear operator $\mathcal{B}$ such that $\mathcal{A}\mathcal{B}=\mathcal{B}\mathcal{A}$ and $U$ is the kernel of $\mathcal{B}$ ?

My thought

First I wanted to find a complementary $\mathcal{A}$-invariant subspace $U'$ such that $V= U \oplus U'$. If this can be done it's easy to solve this problem.

But I found this idea infeasible, and from this post I realized it is a property when the minimal polynomial of $\mathcal{A}$ is a product of distinct irreducibles.

Then I wanted to proceed by assuming that $\mathcal{A}$ has a matrix representation $A=\begin{pmatrix} A_1 & A_2 \\ 0 & A_3 \end{pmatrix}$ under the basis $(\alpha_1,\alpha_2,\dots,\alpha_n)$ where $(\alpha_1,\alpha_2,\dots,\alpha_r)$ is the basis of $U$.

Write $B=\begin{pmatrix} 0 & B_1 \\ 0 & B_2 \end{pmatrix}$. From $AB=BA$ we get \begin{alignat}{2} A_3B_2 & = B_2A_3 \\ A_2B_2 & = B_1A_3-A_1B_1 \\ \end{alignat}

I've known that (supposing we are working on $\mathbb{C}$) $$AX=XB$$ only has zero solution iff $A$ and $B$ have no common eigenvalue.

Supposing we are working on $\mathbb{C}$, then by root subspace decomposition, we can assume that $A_3$ and $A_1$ have common eigenvalues, otherwise we can write $U$ as a union of some root subspaces. (Write $f(\lambda)=\prod_{i=1}^s (\lambda - \lambda_i)^{k_i}$ and $R_i = \ker (\mathcal{A}-\lambda_i\mathcal{I})^{k_i}$. Note that the dimension of the root subspace is its algebraic multiplicity, and hence if $U\cap R_i \ne R_i$, $A_3$ must has an eigenvalue $\lambda_i$. )

Thus we can put $B_2 = 0$ and $B_1 $ being the non-zero solution of $A_1X=XA_3$.

But how do I guarantee that $B_2$ is of full rank? Otherwise it may still not satisfy that the kernel of $\mathcal{B}$ is $U$.

Any hints? Thanks in advance!


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