Let $\mathbb{V}$ be a vector space and $C$ be the matrix associated to a bilinear form $f$ on $\mathbb{V}$ with respect to some choice of initial basis. Show that if the dimension $n$ of $\mathbb{V}$ is odd then there is no basis of $\mathbb{V}$ such that the matrix of $f$ under this new basis is $-C$.
Let the initial basis be $\{ e_1,...,e_n \}$ and the new one $\{ x_1,...,x_n\}$. The entries of the $C$ matrix are $c_{ij}$ which by definition are $f(e_i,e_j)$. By hypothesis also $f(x_i,x_j) = -c_{ij}$. I understand that we must show that there is no transition matrix $A$ such that $-C = A^t C A$, where $A^t$ denotes as usual the transpose of $A$.
How does this matrix looks like given that $f(e_i,e_j)= -f(x_i,x_j)$? I guess that once we know this the exercise will be easy. Thanks in advance.