Do all bilinear forms have a signature? If yes: starting from any arbitrary bilinear form, what is the algorithm to calculate its signature?
If no: what are the conditions necessary for a bilinear form in order to have a signature? Is there an algorithm to calculate a signature for all the bilinear forms that admit one?
 A: Let $b: V\times V\to K$ be a finite-dimensional symmetric bilinear form over some field $K$ (let us say of characteristic not $2$). Then it can be diagonalized, meaning that there is a basis $(e_i)$ of $V$ such that the matrix $(b(e_i,e_j))_{i,j}\in M_n(K)$ is diagonal. If $a_i=b(e_i,e_i)$, we then write
$$b \simeq \langle a_1,\dots,a_n\rangle. $$
The $a_i$ in this diagonalization are absolutely not unique, not even up to reordering: at the very least, you can multiply any $a_i$ by some non-zero square in $K$. On the other hand, there is somethign which is well-defined: the number of $a_i$ which are equal to $0$. That is the dimension of the radical of $b$, and it does not depend on the choice of diagonalization. If that dimension happens to be $0$, we say that $b$ is non-degenerate.
When $K$ is an ordered field, we can say more: the number of $a_i$ which are positive (respectively negative) is also well-defined, independently of the choice of diagonalization. Then the triplet $(r,p,q)\in \mathbb{N}^3$ where $r$ is the number of $a_i$ which are $0$, and $p$ (resp. $q$) is the number of $a_i>0$ (resp. $a_i<0$) does not depend on the diagonalization, and is (often) called the signature of $b$. It always exists when $K$ is an oredered field, and can be computed by diagonalizing $b$ (which can be done in a number of ways, there are probably about 5000 questions on this website on the subject), and then checking the signs of the $a_i$.
Note that the precise terminology might differ depending on the author, for instance what I usually call the signature of $b$ is rather the number $p-q\in \mathbb{Z}$.
Also note that if $K$ is just a field, with no distinguished order, what you can do is define the signature of $b$ at each ordering of $K$ (there might be none of course, that happens precisely when $-1$ is a sum of squares in $K$).
