1
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Yes! I think co-invariant is also used here in exactly the same sense. $\endgroup$ Commented May 17, 2022 at 18:49
2
$\begingroup$

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

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .