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Show that the subspace U is ϕ^z-invariant

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,739
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)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,697
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 $U$.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
So in general the statement of the exercise is not true, only if we consider some restrictions? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,739
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 $U\le_{\text{id}} V$.