Define $f: V \times V \rightarrow \mathbb{K}$ ($\mathbb{K}$ field), by $f(u, v) = \langle u, Av\rangle$

Question

Define $f: V \times V \rightarrow \mathbb{K}$ ($\mathbb{K}$ field), by $f(u, v) = \langle u, Av\rangle$, any $u$, $v \in V$ and $A \in M(n, \mathbb{R})$ with $A^T = -A$, i.e., $A$ is skew-symmetric.

a) Prove $f$ is skew-symmetric.

b) Prove $f$ is symplectic if and only if $A$ is invertible.

c) Prove the item b) implies which $n$ must be even. If you can help me.

The item a), founded is:

$f$ is skew-symmetric if $f(u,v) = - f(v,u)$. Then $f(u,v) = \langle u, Av\rangle = \langle A^T u, v\rangle = \langle -Au, v\rangle = - \langle v, Au\rangle = - f(v,u)$.

But I don't understand the first step: Why $\langle u, Av\rangle = \langle A^T u, v\rangle$ ?

Help me, please.

Note: for b), $f$ is symplectic if $f$ is bilinear form with the propriety $f(u,v) = -f(v,u)$ and $\ker f = \{0\}$, where $\ker f = \{v \in V; f(u,v) = 0, \mbox{for any}\; u \in V\}$.

Note2: For matrices in bilinear forms, the only thing I saw is that if $V$ is a vector space of $n$ finite dimension, we can consider $f: V \times V \rightarrow \mathbb{K}$, with $f: \mathbb{K}^n \times \mathbb{K}^n \rightarrow \mathbb{K}$, by $(e_i, e_j) \mapsto f(e_i, e_j) = B_{ij}$ and then consider $\{e_1, e_2, \ldots, e_n\}$ basis of $\mathbb{K}^n$, $u = \sum u_ie_i$, $v=\sum v_je_j$, we obtain for a general bilinear form $f(u,v) = \displaystyle\sum_{i,j=1}^{n}u_iv_jB_{ij}$. But I can not understand the items b) and c) what I mentioned in the item a). Sorry if the question is very simple to you, but I found in search and can not solve it.

Answer 1

For (a) we wish to show that $$ \langle u,Av\rangle = \langle A^\top u,v\rangle $$ To do so, we compute \begin{align*} \langle u,Av\rangle &= \sum_{i=1}^n u_i[Av]_i \\ &= \sum_{i=1}^n u_i\sum_{j=1}^n A_{ij}v_j \\ &= \sum_{j=1}^nv_j\sum_{i=1}^nA_{ij}u_i \\ &= \sum_{j=1}^n v_j\sum_{i=1}^n A^\top_{ji}u_i \\ &= \sum_{j=1}^n v_j[A^\top u]_j \\ &= \sum_{j=1}^n [A^\top u]_j v_j \\ &= \langle A^\top u,v\rangle \end{align*}

Addendum: If we trust the formula $$ \langle x,y\rangle=x^\top y $$ then our identity is quite obvious. We have $$ \langle u, A\,v\rangle= u^\top (A\,v)=\left((A\,v)^\top u\right)^\top =\left((v^\top A^\top)u\right)^\top=(v^\top (A^\top\, u))^\top=(A^\top\,u)^\top v=\langle A^\top u,v\rangle $$