Prove that if $\phi$ is invertible operator of V space, then $\phi$ and $\phi^{-1}$ has the same invariant subspaces.

Question

Prove that if $\phi$ is invertible operator of V space, then $\phi$ and $\phi^{-1}$ has the same invariant subspaces. I know how to prove this statement for finite V. (Wish it is right) : Let $W$ be invariant subspace for $\phi$, then $\phi(W)\subset W$. If W is finite and $\phi$ is invertible $\Rightarrow$ $\phi\mid_W : W \rightarrow W$ is bijection, that's why for every $v\in W$ there exist $v'\in W$ such that $\phi(v) = v' \Rightarrow \phi^{-1}(\phi(v)) = \phi^{-1}(v')\Rightarrow v = \phi^{-1}(v') \Rightarrow \phi^{-1}(W)\subset W$.

But I don't know how to deal with infinite V. Can I do the same or it is completely different solution? Can you help me?

Answer 1

I suppose that the assertion that you wish to prove is this:

If $V$ is a vector space, $\phi\colon V\longrightarrow V$ is an invertible linear map and $W\subset V$ is a subspace such that $\phi(W)\subset W$, then $\phi^{-1}(W)\subset W$.

If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $\mathbb R$ into itself, let $\bigl(\phi(f)\bigr)(x)=f(x-1)$ and let $W=\{f\colon\mathbb{R}\longrightarrow\mathbb{R}\,|\,x<0\implies f(x)=0\}$. Then $\phi(W)\subset W$. However, if you define$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<0\\1&\text{ otherwise,}\end{cases}\end{array}$$then $f\in W$, but $\phi^{-1}(f)\notin W$, since $\bigl(\phi^{-1}(f)\bigr)\left(-\frac12\right)=1$.