What do you call the converse of an invariant subspace of an operator?

Question

Question. I am looking for the concept converse to invariance: what do we call a set $W$, such that $$T(w) \in W \implies w \in W ?\tag{1}$$

I feel there was a word for this, but I can't recall it, maybe something related to "absorbent" or $\alpha$-limit... A suitable adjective, noun, a convenient phrase (or an explanation why they don't exist) count as a good answer.

Context. $W\subseteq V$ is called an invariant set or subspace of a linear operator $T:V\to V$ if $$T(W)\subseteq W.$$ Equivalently, $W$ is called $T$-invariant if $w \in W \implies T(w) \in W$.

For simplicity, assume that $\dim V = n < \infty$ and $T\in\mathbb{R}^{n\times n}$ is a matrix. If $T$ is invertible, then $$T(w) \in W \implies w \in W,$$ becomes, after substituting $w=T^{-1}(v)$, $$v \in W \implies T^{-1}(v) \in W, \tag{4}$$ which is equivalent to saying $ W \text{ is } T^{-1}\text{-invariant}$ and writing $$T^{-1}(W) \subseteq W.\tag{5}$$

If $T$ is not invertible, we can still use (4) and (5) by defining $T^{-1}$ as the pre-image of $T$. If I am not mistaken, this is also equivalent to a linear (e.g. orthogonal) complement of $W$ being $T$-invariant, $$ T(W^{\perp})\subseteq W^{\perp}.$$ However, I find all these characterisations awkward and am looking for a better and more expressive word.

This question came up when I was thinking about Schur decomposition, Jordan normal form and generalised eigenspaces. An eigenspace is an invariant subspace in which $T$ is "spherical". Thinking about the invariance of eigenspaces does not get us far, because the invariance does not address generalised eigenvectors (as defined here). Similarly to a regular eigenspace, a generalised eigenspace is $T$-invariant, that, is we cannot leave it. But importantly, we cannot enter it either.

Answer 1

By taking the contrapositive, this is equivalent to $w \in W^{C} \to T(w) \in W^{C}$, so $W^{C}$ is $T$-invariant. Perhaps $W$ should then be called co-invariant?

Answer 2

Elchanan Solomon's lightning-fast answer is acknowledged. However, more can be said.

A subspace $\mathcal{M}\subset \mathbb{C}^n $ is called coinvariant for the transformation $A: \mathbb{C}^n \to \mathbb{C}^n$: if there is an $A$-invariant direct complement to $\mathcal{M}$ in $\mathbb{C}^n $.

A subspace $\mathcal{M}\subset \mathbb{C}^n $ is called orthogonally coinvariant for the transformation $A: \mathbb{C}^n \to \mathbb{C}^n$: if the orthogonal complement $\mathcal{M}^\perp$ to $\mathcal{M}$ is $A$-invariant.

These definitions are found on pp 105 and 108 of

• Israel Gohberg, Peter Lancaster, and Leiba Rodman. Invariant Subspaces of Matrices with Applications. SIAM (Classics in applied mathematics), 2006. https://doi.org/10.1137/1.9780898719093

As the title suggests, the book is an awesome source on this topic and includes clear examples and relevant results. Here is a close quote of two of them, which illustrate the relation to the adjoint $A^*$:

Proposition 3.1.2. A subspace $\mathcal{M}$ is $A$-coinvariant iff its orthogonal complement $\mathcal{M}^\perp$ is $A^*$-coinvariant.

Proposition 3.1.3. A subspace $\mathcal{M}$ is orthogonally $A$-coinvariant iff it is $A^*$-invariant.