# invariant subspace of the inverse

Question: Prove that if $T:V \to V$ is a linear map, which is invertible, then each invariant subspace of T is also invariant under $T^{-1}$. V is finite.

Thoughts I thought about composing them and somehow getting the identity map (which is also included in the invariant subspace if multiplied by some vector w), But this doesn't seem too formal..

• That is not (necessarily) true for infinite-dimensional spaces. Is $V$ assumed to be finite-dimensional? – Daniel Fischer Feb 11 '14 at 21:05
• As stated, the claim is not true. Let $V$ be the space of ufunctions $f\colon\mathbb Z\to\mathbb R$ and $Tf(k)=f(k+1)$. Then $\{f:f|_{\mathbb N}=0\}$ is invarinat under $T$ but not under $T^{-1}$. – Hagen von Eitzen Feb 11 '14 at 21:06
• forgot to mention V was finite. – jreing Feb 11 '14 at 21:09

Let $V$ be an invariant subspace under $T$. This means $T(V) \subseteq V$.
Now, if $V$ is finite dimensional, and since $T$ is invertible, you can easily show that $dim(T(V))=dim(V)$ and hence we actually have $T(V)=V$
Now, let $v \in V$. Then $v=T(w)$ for some $w \in V$. Hence, $T^{-1}(V) \subseteq V.$
If $W\subseteq V$ is $T$-invariant, then $T$ restricts to a linear map $T\vert_W:W\rightarrow W$. This map is injective since $T$ is injective. So, $T$ maps $W$ isomorphically onto $W$. Hence, $T^{-1}$ maps $W$ isomorphically onto $W$.