Bilinear forms and congruent matrices

Question

Let $V$ be a finite dimensional vector space over $\mathbb{K}$ and $\phi, \psi: V \times V \rightarrow \mathbb{K}$ be two bilinear forms.

I'm investigating the relation between the follwing two propositions:

1. There exist an invertible linear operator $L$ on $V$ such that $\psi(Lv,Lw)=\phi(v,w)$ $\forall v,w \in V$

2. $\forall$ $ \mathcal{V}$ basis of $V$ the two representing matrices $A_{\phi}^{\mathcal{V}}$, $A_{\psi}^{\mathcal{V}}$ are congruent

This is my work:

$(1) \implies (2): $ choose a basis $\mathcal{V}$ of $V$. Suppose $P_L^{\mathcal{V}}$ represents $L$ with respect to $\mathcal{V}$. Let $[v]_{\mathcal{V}}$ denote the vector of coordinates of $v \in V$ in $\mathcal{V}$. Then $\psi(Lv,Lw)=[Lv]_{\mathcal{V}}^T A_{\psi}^{\mathcal{V}} [Lw]_\mathcal{V}=[v]_{\mathcal{V}}^T P_{\mathcal{V}}^T A_{\psi}^{\mathcal{V}} P_{\mathcal{V}}[w]_{\mathcal{V}}= [v]_{\mathcal{V}}^T A_{\phi}^{\mathcal{V}} [w]_{\mathcal{V}}=\phi(v,w)$ for every $v,w \in V$.

$(2) \implies (1)$: I know that $\forall$ $\mathcal{V}$ basis of $V$ there's an invertible matrix $P_{\mathcal{V}}$ that may depend on $\mathcal{V}$ such that $P_{\mathcal{V}}^T A_{\psi}^{\mathcal{V}} P_{\mathcal{V}}= A_{\phi}^{\mathcal{V}}$. Calling $L_{\mathcal{V}}$ the linear operator associated to $P_{\mathcal{V}}$ one can conclude the following:

$\forall$ $\mathcal{V}$ basis of $V$ there's an invertible linear operator $L_{\mathcal{V}}$ such that $\psi(L_{\mathcal{V}}v,L_{\mathcal{V}}w)=\phi(v,w)$ $\forall v,w \in V$

Do you think I can get rid of the dependence of $\mathcal{V}$ in $L$ to obtain the second proposition? Any hint or counterexample is appreciated!

Answer 1

To approach $(2) \Rightarrow (1)$ we can prove that $(2) \Rightarrow (3):= \exists B,B'$ basis of $V$ : $M_{B}(\phi) = M_{B'}(\psi)$ and then that $(3) \Rightarrow (1)$. This should solve the problem of $\mathcal{V}$

$(2) \Rightarrow (3)$ Let's fix $B$ basis of $V$. We know that $M_{B}(\phi) = P^{\top}M_{B}(\psi)P, \hspace{0.2cm} P \in GL(V)$, where $GL(V)$ denotes the group of the invertible endomorphisms of V.

Besides, from change of basis theory we know that $\exists! B'$ such that $M_{B', B}(id) = P$

In the basis $B'$ we have :

$$x^{\top}M_{B'}(\psi)y = (Px)^{\top}M_{B}(\psi)(Py) \hspace{0.2cm} \forall x,y \in \mathbb{K}^{n}$$

Where $x,y$ are thought as vectors in the basis $B'$.

Hence $$M_{B'}(\psi) = P^{\top}M_{B}(\psi)P \Rightarrow M_{B}(\phi) = M_{B'}(\psi) \hspace{0.2cm}\Box.$$

Now we prove that $(3) \Rightarrow (1)$ :

Similarly, let $P=M_{B',B}(id)$. Obviously $P \in GL(V)$.

It is true that $P$ is really what we're looking for, or in other words isometry between $(V,\phi), (V,\psi)$

(If thought as matrix of the endormorphism $f$ that maps $B\to B'$),

In fact, calling $[]_{B}$ the isomorpshism of coordinates $\forall v,w \in V, [v]_{B}=x,[w]_{B}=y$ we have :

$$\phi(v,w) = x^{\top}M_{B}(\phi)y = x^{\top}M_{B'}(\psi)y = x^{\top}P^{\top}M_{B}(\psi)Py = \psi(f(v),f(w)) \hspace{0.2cm} \Box.$$

(If $P$ is thought as matrix of $f$).

I think this answers your question, let me know if there is anything that doesn't feel right to you.