Congruent matrices - why do we require invertiblility?

Question

If $A$, $B$ $\in K^{n \times n}$ are $n \times n$ matrices over a field $K$, then we say that $A$ and $B$ are congruent if there exists an invertible $P \in GL(n, K)$ such that $B = P^TAP$, where $P^T$ denotes the transpose of $P$.

Why do we require $P$ to be invertible?

Congruence of matrices is usually compared to similarity of matrices, where we say that $A$ and $B$ are similar if there exists an invertible $P \in GL(n, K)$ such that $B = P^{-1}AP$. Here it is obvious why we need $P$ to be invertible. However, in the case of congruent matrices, the invertibility requirement doesn't seem to be obvious.

Answer 1

We need that congruence is an equivalence relation. If we have $A\sim B$ iff $B=P^tAP$ for some $P$, then reflexivity says that also $B\sim A$ should hold, i.e., $A=Q^tBQ$ for some $Q$. If $P$ is invertible, this follows easily. If not, we have a problem. For example, for every matrix $A$ we would have $A\sim 0$ with $P=0$, but $0\sim A$ would force $A=0$, which is not true for all $A$.

Answer 2

The word "congruent" (and the fact that you write "$A$ and $B$ are congruent", rather than "$A$ is congruent to $B$") suggests an equivalence relation. In particular, in order for this relation to be symmetric, you will need $P$ to be invertible in general.

Of course, you could take the equivalence relation generated by the relation you suggest, but this is the total relation, as then any matrix will be "congruent" to the zero matrix.