Suppose $V, V'$ are $n$-dimensional vector spaces over $\mathbb C$, with symmetric bilinear forms $ \langle \cdot, \cdot \rangle$ and $\langle \cdot, \cdot \rangle'$ respectively.
I am asked to show that $\mathfrak{so}(V) \cong \mathfrak{so}(V')$, i.e. that the Lie algebra $\mathfrak{so}_n(\mathbb C)$ is independent of the choice of symmetric bilinear form.
Here's my thinking so far:
We may find matrices $M \in GL(v), \; M' \in GL(V')$ to represent the two above symmetric bilinear forms respectively. Thus:
$\mathfrak{so}(V) = \{x \in \mathfrak{gl}(V) \mid Mx + x^TM = 0\}, \; \mathfrak{so}(V')= \{x' \in \mathfrak{gl}(V') \mid M'x' + (x')^TM' = 0\}$
Now I have to find a homomorphism of Lie algebras $\phi : \mathfrak{so}(V) \rightarrow \mathfrak{so}(V')$ that is also an isomorphism of the vector spaces $\mathfrak{so}(V)$ and $\mathfrak{so}(V')$.
Now, $\mathfrak{gl}(V)$ and $\mathfrak{gl}(V')$ contain the same matrices, that is they are both the set of $n \times n$ matrices over $\mathbb C$. Given this, I am thinking of starting with just the "identity" map from $\mathfrak{gl}(V)$ to $\mathfrak{gl}(V')$ where we have:
$i : \mathfrak{gl}(V) \rightarrow \mathfrak{gl}(V')$ such that:
$i(M) = M'$ where $M, M'$ have the same entries, but belond to the "different" vector spaces $\mathfrak{gl}(V)$ and $\mathfrak{gl}(V')$.
Clearly (?) this is an isomorphism of vector spaces for $\mathfrak{gl}(V)$ and $\mathfrak{gl}(V')$, but does it restrict to an isomorphism between $\mathfrak{so}(V)$ and $\mathfrak{so}(V')$?
Around this point I get stuck and I don't know how to proceed. I'm fairly new to Lie algebras and aren't comfortable using them yet, so I can't quite see what my next step should be. Any help would be greatly appreciated, thank you!