Proof verification : T-invariant subspace has T-invariant complement

Question

I was asked to prove the following Lemma :

Lemma: Let $\varphi$ be a symmetric or alternating bilinear form , V a finite-dimensional space , T an isomorphism $T:V\rightarrow V$ such that $\varphi((T(u),T(v)) = \varphi (u,v) , \forall u,v \in V$ and W a T- inavariant non- degenerated subspace of V ( $W\cap W^{\perp} = 0$). Then $W^{\perp}$ is T-invariant .

Proof: Let $u \in W^{\perp} $. Then $\varphi (u,w)= 0 , \forall w\in W$ $\Rightarrow \varphi (u,w)= \varphi((T(u),T(w)) = 0$ .Because W is T-invariant ,$T(W) \subseteq W$ .But T is an isomorphism $\Rightarrow T(W) = W$. So we have that :

$0 = \varphi((T(u),z) , \forall z \in W = T(W)$ $\Rightarrow T(u )\in W^{\perp}$ $\Rightarrow W^{\perp}$ is T-invariant.

My doubt is: I didn't use the hypothesis that W is non degenerated. So did I miss something ? Is this hypothesis really necessary ?

Answer 1

Let us assume that all vector spaces are finite-dimensional (I think the infinite-dimensional case could cause some problems).

As you have shown in your proof (which is correct except for the minor details I mentioned), the assumption that $W$ is a non-degenerate subspace is not necessary to show that $W^\perp$ is $T$-invariant.

Only if you want to show that $\require{enclose} \enclose{horizontalstrike}{T(W^\perp) = W^\perp}$, then we must assume that $W$ is non-degenerate.

Edit: The remark I crossed out is wrong. We always get $T(W^\perp) = W^\perp$, regardless of $W$ being non-degenerate or not. The reason is that $T(W^\perp) \subseteq W^\perp$ (as you have shown in your proof) and T is an isomorphism, so $T(W^\perp) = W^\perp$. The argument is exactly the same which you used to show that $T(W)=W$. Sorry once again!