Let $V$ be a finite-dimensional vector space, and let $M: V \rightarrow V$ be a linear transformation. Suppose subspaces $S_1,S_2 \subset V$ are both $M$-invariant and have $M$-invariant complements (say $W_1$ and $W_2$, so $V = S_i \oplus W_i$).
Question: do the sum $S_1 + S_2$ and intersection $S_1 \cap S_2$ also have $M$-invariant complements?
I haven't yet found a valid proof or counterexample, and would appreciate any intuition (or proof or counterexample, of course) that anyone has towards this question. I know that the answer is yes if $M$ is diagonalizable or if the $W_i$ are actually the orthogonal complements. Can we do it without placing those restrictions?
Thanks!
(Here's what I've tried so far.)
My guess is that the answer is yes, and my proof attempts are inspired primarily by Theorem 2.5.1 of this book.
The idea is to split $S_1$ and $S_2$ into irreducible subspaces (where an "irreducible subspace" is an $M$-invariant subspace that cannot be written as the direct sum of other $M$-invariant subspaces).
Before starting, also define a "supremal irreducible subspace" to be an irreducible subspace that is not a subset of any other irreducible subspace.
With that, here are the broad steps:
-
I've shown that an invariant subspace has an invariant complement if and only if it is the direct sum of supremal irreducible subspaces.
So the subspaces $S_1$ and $S_2$ can both be written as direct sums of supremal irreducible subspaces. -
Since $S_1$ and $S_2$ are both $M$-invariant, so are $S_1 + S_2$ and $S_1 \cap S_2$. Those sum and intersection can be written as direct sums of irreducible subspaces (by definition), but those irreducible subspaces are not necessarily supremal.
-
If we can prove that those irreducible subspaces are supremal, then we're done… but I haven't had any luck with that so far. It's clearly dependent on choosing decompositions for $S_1$ and $S_2$ that fully characterize the supremal irreducible subspaces common to both of them.
-
A first step might be to show that any two supremal irreducible subspaces are either equal or independent, and I think I can do the rest. Equivalently, we can show that if two Jordan chains have the same eigenvector and the same length, then their spans are equal.
Best Answer
False! Counterexample: $V = \mathbb{R}^4$ and $$ M = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \,. $$ Take the two $M$-invariant subspaces $S_i$ (and $M$-invariant complements $W_i$) $$ S_1 = \mathrm{span} \left\{ \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix} \right\} \,,\ S_2 = \mathrm{span} \left\{ \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\1\\1\\0 \end{bmatrix} \right\} \,,\quad W_1 = W_2 = \mathrm{span} \left\{ \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\1 \end{bmatrix} \right\} \,. $$ The intersection $S_1 \cap S_2$ has only part of a Jordan chain for the top Jordan block of $M$, so it can't have an $M$-invariant complement. The sum $S_1 + S_2$ has only part of a Jordan chain for the bottom Jordan block of $M$, so it also can't have an $M$-invariant complement.