Quadratic forms -- rank of matrix Assume that $M$ is the matrix of some quadratic form (over any field of characteristic not $2$) and set
$$Q(\overline{x})=\overline{x}^tM\overline{x}$$
We can replace $M$ by the symmetric matrix $N=\frac{1}{2}(M^t+M)$ and $N$ defines the same form, i.e.
$$\overline{x}^tN\overline{x}=\overline{x}^tM\overline{x}$$
for all choices of the vector $\overline{x}$. I would assume that this matrix $N$ has the same rank as the matrix $M$, but I can't seem to be able to prove it. In other words, is this true and how can I prove it? I would assume that the rank of the form is some intrinsic property that can be defined without referring to the matrix used.
Books I've looked at seem to assume that this substitution has been made a priori, so this question is never answered.

No, the rank is not the same.  For example,
$ M = \pmatrix{0 & 1\cr -1 & 0\cr}$ has rank $2$, but $(M'+M)/2$ has rank $0$. 
