# An invariant subspace of $\mathcal{A}$ must be the kernel of a commuting linear operator $\mathcal{B}$

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 $$U$$.