Relation between vector space basis and bilinear form linked to Lie algebra

I have a bilinear form $$\phi$$ on a complex vector space V and I have to prove the following:

1) if $$\phi$$ is symmetric non-degenerate it is possible to choose a basis for V such that $$\phi$$ is described by $$X^TY$$. Deduce that special linear Lie algebra $$\mathfrak{so}(V,\phi)\simeq \mathfrak{so_n}$$

where $$\mathfrak{so}(V,\phi)= \{A \in M_{n,n}(\mathbb{C}) : (A(u),v)+(u,A(v))=0, \forall u,v \in V\}$$

2) if $$\phi$$ is skew-symmetric non-degenerate it is possible to choose a basis for V such that $$\phi$$ is described by $$X^TJY$$, where $$J=\begin{bmatrix} 0 & \mathbb{I}\\ -\mathbb{I} & 0\\\end{bmatrix}$$

I know that the matrix $$B$$ associated to $$\phi$$ must have $$\det B \neq 0$$, but how to relate this with the basis? Why there is a difference in the basis for symmetric and skew-symmetric bilinear forms?

I am struggling with this exercise, can anyone help me?

Define $$U := \{u \in V: J(u,v) = v \text{ for any } v \in V \}$$. Since $$J$$ is non-degenerate, then $$U = \{0\}$$. So, if $$V$$ is not trivial, there exists an element $$e_1 \in V \setminus \{0\}$$ such that $$J(e_1,f_1) \neq 0$$ for some $$f_1 \in V$$. Up to rescaling, you can assume $$J(e_1,f_1) = 1$$. Call $$W := \text{Span}(e_1,f_1)$$ and define $$W^J := \{u \in V: J(u,w) = 0 \text{ for any }w \in W\}.$$ Let us take a look at $$W \cap W^J$$. If $$v \in W \cap W^J$$, then $$v = ae_1+bf_1$$ for some $$a,b$$, and $$J(v,e_1) = 0 = J(v,f_1)$$. But then $$J(v,e_1)=J(ae_1+bf_1,e_1) = -b=0$$ and similarly $$a=0$$. So $$W \cap W^J = \{0\}$$. Let now $$v$$ be any vector in $$V$$. If $$J(v,e_1) = -a$$ and $$J(v,f_1) = b$$, then you can write $$v = be_1+af_1+v-be_1-af_1.$$ You have that $$be_1+af_1 \in W$$ and \begin{align} J(v-be_1-af_1,e_1) & = J(v,e_1)+a = a-a=0\\ J(v-be_1-af_1,f_1) & = J(v,f_1)-b = b-b = 0. \end{align} This tells you that any vector $$v \in V$$ can be written as a sum of a vector in $$W$$, that is $$be_1+af_1$$, and a vector in $$W^J$$, namely $$v-be_1-af_1$$. Consequently $$V = W \oplus W^J$$. If $$W^J = \{0\}$$, then you are done and $$J$$ can be written in the basis $$\{e_1,f_1\}$$ as $$\left( \begin{matrix} J(e_1,e_1) & J(e_1,f_1) \\ J(f_1,e_1) & J(f_1,f_1) \end{matrix} \right) = \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right).$$ Otherwise, choose $$e_2 \neq 0$$ in $$W^J$$ and repeat the process getting $$f_2$$ such that $$J(e_2,f_2)=1$$. Going on you will find a basis $$\{e_1,e_2,\dots,e_n,f_1,f_2,\dots,f_n\}$$ of $$V$$ such that $$J(e_i,e_j) = 0, \quad J(e_i,f_k) = \delta_{ik}, \quad J(f_i,f_j) = 0,$$ where $$\delta_{ij}$$ denotes the Kronecker delta. The process ends after $$n$$ steps, as $$\dim V < \infty$$. Notice that the presence of $$J$$ forces $$\dim V = 2n$$, i.e. the dimension of $$V$$ is even.
FYI: a non-degenerate skew-symmetric bilinear form $$J$$ like this is generally called symplectic form on $$V$$, and $$(V,J)$$ is then called symplectic space.
• Thanks! Can I ask you another question? Why $(A(u),v)+(u,A(v)) = A^T J+ J A ?$ – Phi_24 Oct 2 '18 at 19:56
• Take a look here. Just consider that your $J$ is $\Omega$. – Gibbs Oct 2 '18 at 20:02