Show that the subspace U is ϕ^z-invariant

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Subspace
In summary, the conversation discusses the idea of showing that if $\phi$ is an invertible linear map and $U$ is a subspace of $V$ that is $\phi$-invariant, then it is also $\phi^z$-invariant for all $z\in\mathbb{Z}$, if certain restrictions are considered. The statement is true for finite-dimensional vector spaces, but not for infinite-dimensional ones. It is proven using the fact that the images of a basis under an invertible linear map are also a basis, and the trivial case for $z=0$.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! 😊

Let $\mathbb{K}$ be a field and let $V$ be a $\mathbb{K}$-vector space.

Let $\phi,\psi:V\rightarrow V$ be linear maps, such that $\phi\circ\psi=\psi\circ\phi$.

I have shown using induction that if $U\leq_{\phi}V$ (i.e. it $U$ is a subspace and $\phi$-invariant), then $U\leq_{\phi^k}V$ for all $k\in \mathbb{N}$.

Now I want to show that if $\phi$ is invertible and if $U\leq_{\phi}V$, then $U\leq_{\phi^z}V$ for all $z\in \mathbb{Z}$. My idea is the following:

If $z=:n\in \mathbb{Z}_{> 0}$, so $z=n\in \mathbb{N}$, the from the previous result it follows that $U$ $\ \phi^n$-invariant.

Since $\phi$ isinvertible there is a linear map $\chi$ such that $\chi:=\phi^{-1}$.

If $z=:-n\in \mathbb{Z}_{< 0}$, with $n\in \mathbb{N}$, then $\phi^{-n}=\left (\phi^{-1}\right )^n=\chi^n$. Then if we show that $U$ is $\chi$-invariant then it follows from the previous result that $U$ is $\ \phi^{-n}$-invariant. Is that correct?

Or do we show that in an other way?

:unsure:
 
Physics news on Phys.org
  • #2
mathmari said:
Let $\mathbb{K}$ be a field and let $V$ be a $\mathbb{K}$-vector space.

Let $\phi,\psi:V\rightarrow V$ be linear maps, such that $\phi\circ\psi=\psi\circ\phi$.

I have shown using induction that if $U\leq_{\phi}V$ (i.e. it $U$ is a subspace and $\phi$-invariant), then $U\leq_{\phi^k}V$ for all $k\in \mathbb{N}$.

Now I want to show that if $\phi$ is invertible and if $U\leq_{\phi}V$, then $U\leq_{\phi^z}V$ for all $z\in \mathbb{Z}$.My idea is the following:

If $z=:n\in \mathbb{Z}_{> 0}$, so $z=n\in \mathbb{N}$, the from the previous result it follows that $U$ $\ \phi^n$-invariant.

Since $\phi$ isinvertible there is a linear map $\chi$ such that $\chi:=\phi^{-1}$.

If $z=:-n\in \mathbb{Z}_{< 0}$, with $n\in \mathbb{N}$, then $\phi^{-n}=\left (\phi^{-1}\right )^n=\chi^n$. Then if we show that $U$ is $\chi$-invariant then it follows from the previous result that $U$ is $\ \phi^{-n}$-invariant.Is that correct?

Or do we show that in an other way?
Hey mathmari!

I don't think it is true. 😢

Counterexample
Consider $\phi=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\psi=I_2$, $U=\left\langle e_1\right\rangle$, and $V=\mathbb R^2$.
We have $U\le_\phi V$, but we don't have $U\le_{\phi^{-1}}V$ do we? (Worried)
 
  • #3
Klaas van Aarsen said:
Hey mathmari!

I don't think it is true. 😢

Counterexample
Consider $\phi=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\psi=I_2$, $U=\left\langle e_1\right\rangle$, and $V=\mathbb R^2$.
We have $U\le_\phi V$, but we don't have $U\le_{\phi^{-1}}V$ do we? (Worried)
But that $\phi$ is not invertible. I think that the result is true if $\phi$ is invertible and the space $V$ is finite-dimensional over $\Bbb K$, but not if $V$ is infinite-dimensional.

For example, if $V$ has a basis $\{e_n:n\in\Bbb{Z}\}$, $U$ is the subspace spanned by $\{e_n:n\geqslant0\}$ and $\phi$ is the shift map given by $\phi(e_n) = \phi_{n+1}$, then $U$ is invariant under $\phi$, but not under the inverse map $\phi^{-1}$, which is the backward shift map taking $e_0$ to $e_{-1}$ which is not in $U$.
 
  • #4
So in general the statement of the exercise is not true, only if we consider some restrictions? :unsure:
 
  • #5
mathmari said:
So in general the statement of the exercise is not true, only if we consider some restrictions?

If $V$ is allowed to be infinite-dimensional, it is not generally true as Opalg pointed out.

Still, if $V$ is finite-dimensional, we can prove it.

Since $U\le_\phi V$, we have that $\phi(U)$ is a linear subspace of $U$.
Let $\{u_i\}$ be a basis of $U$.
Since $\phi$ is invertible, $\{\phi(u_i)\}$ must be a set of independent vectors.
As they must also be in $U$, it follows that $\{\phi(u_i)\}$ is a basis of $U$ as well.
Therefore $\phi^{-1}(u_i)$ are all in $U$ and we have that $U\le_{\phi^{-1}} V$. :geek:

The result for $\phi^{-n}$ follows as you already found. (Nod)

That leaves the case $z=0$, which is trivial since $U\le_{\text{id}} V$.
 

1. What does it mean for a subspace to be ϕ^z-invariant?

When a subspace U is ϕ^z-invariant, it means that for any vector v in U, applying the linear transformation ϕ^z to v will result in another vector that is also in U.

2. How do you show that a subspace is ϕ^z-invariant?

To show that a subspace U is ϕ^z-invariant, you need to prove that for any vector v in U, ϕ^z(v) is also in U. This can be done by using the properties of linear transformations and the definition of a subspace.

3. Can a subspace be ϕ^z-invariant for some vectors but not others?

Yes, it is possible for a subspace to be ϕ^z-invariant for some vectors but not others. This depends on the specific properties of the linear transformation ϕ^z and the subspace U.

4. Why is it important to show that a subspace is ϕ^z-invariant?

Showing that a subspace is ϕ^z-invariant can help us understand the behavior of the linear transformation ϕ^z and how it affects vectors in the subspace U. It can also be useful in solving problems involving linear transformations and subspaces.

5. Can a subspace be ϕ^z-invariant for all linear transformations?

No, not all subspaces are ϕ^z-invariant for all linear transformations. The ϕ^z-invariance of a subspace depends on the specific properties and behavior of the linear transformation ϕ^z and the subspace U.

Similar threads

  • Linear and Abstract Algebra
2
Replies
39
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
828
  • Linear and Abstract Algebra
Replies
10
Views
2K
  • Linear and Abstract Algebra
Replies
9
Views
983
  • Linear and Abstract Algebra
Replies
9
Views
972
  • Linear and Abstract Algebra
Replies
12
Views
1K
Replies
10
Views
1K
Replies
24
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
956
Back
Top