- #1

mathmari

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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: