0
$\begingroup$

I just, just, started reading about invariant subspaces, but I don't think I'm getting a really concrete idea of what they are. Could someone try to explain to me more advanced examples of this?

This is what I know so far; If we let $V$ be a the $F$-vector space and let a linear transformation exist that maps onto itself, i.e $T \in L(V)$. Then, a subspace $w$ of $V$ is invariant under $T$ if $T(w) = w$ for all $w \in W$.

I think a very generalized way to say this would be that if you take a linear operator and apply it to a subspace where the subspace doesn't change, then it is invariant.

I get this definition, but how does this apply to a span example? Or an upper triangle example? Or any non-trivial examples, i.e not just $ [0]$?

$\endgroup$
  • $\begingroup$ $T(w) = w$ means that $W$ is fixed under $T$, which is quite different from being invariant under $T$. $\endgroup$ – lhf Jan 29 '15 at 0:08
  • $\begingroup$ $T(w) = w$ means that $W$ is fixed under $T$, which is quite different from being invariant under $T$. $\endgroup$ – lhf Jan 29 '15 at 0:08
  • $\begingroup$ Okay, thank-you for correcting me! I must have written it down wrong, and I see I didn't really understand what I was talking about. I'm glad you told me! So just to clarify, $T(w) \in W$ is the correct version right? $\endgroup$ – user180708 Jan 29 '15 at 0:14
1
$\begingroup$

There's an error in your formulation: a subspace $W$ of $V$ is $T$-invariant if $T(w)\in W$ for all $w\in W$. In other words: $W$ is $T$-invariant if $T(W) \subset W$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Here is a simple example. Consider the action of the symmetric group $S_n$ on $\mathbb{R}^n$: for each $\pi \in S_n$, $(x_1,\ldots,x_n)^\pi = (x_{\pi(1)},\ldots,x_{\pi(n)})$. This action preserves the sum of all coordinates, and so the subspace of constant vectors is invariant, as is its complement, the subspace of all vectors with zero sum.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Subspace of constant vectors? What's that? $\endgroup$ – Rob Arthan Jan 29 '15 at 0:01
  • $\begingroup$ It is the subspace spanned by the vector $(1,\ldots,1)$. $\endgroup$ – Yuval Filmus Jan 29 '15 at 0:15
  • $\begingroup$ I had to look up what a symmetric group is, but that's a great example, thank-you! But wait - the sum has to remain the same? I thought the definition just meant that the same components had to be in it? What if we took the absolute values of components. The sum would be the same, but the components wouldn't, would they? $\endgroup$ – user180708 Jan 29 '15 at 0:17
  • $\begingroup$ The sum is preserved since $x_1+\cdots+x_n = x_{\pi(1)}+\cdots+x_{\pi(n)}$. $\endgroup$ – Yuval Filmus Jan 29 '15 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.