As we known, for matrix similarity $A= P^{-1}B P$, we can classify equivalence class in $M_n (\mathbb{C})$ by Jordan Canonical Form.
Matrix congruence, $A = P^T B P$ with $P$ invertible, is also an equivalence relation. How does one classify the equivalence classes in $M_n(\mathbb{R})$ under this relation?
For symmetric real matrices, it's easy, since we can use orthogonal transformation to diagonalize it, which gives both matrix similarity and matrix congruence. Then we use a scaling matrix to make the diagonal elements to be $\pm 1$ and $0$. Therefore, for a symmetric real matrix, we classify the matrix by the number of $\pm1$ and $0$. What about for general real matrices?
Best Answer
I don't know the answer for real matrices of general dimension (see below for the $n = 2$ case), but for complex matrices the description of the congruence equivalence classes was worked out by Aitken and Turnbull, and there is a nice statement of the classification theorem in $\S\,$3 of the conference paper user1551 linked in the comments: In short, one defines several families of building block matrices, and the theorem asserts that every matrix is congruent to a direct sum of such blocks.
Here's how one can proceed: For a general $n \times n$ matrix $A$, the properties of symmetry and skewness are invariant under congruence, so if we decomposes any $n \times n$ matrix into its symmetric and skew parts, $$A = \underbrace{\tfrac{1}{2}(A + A^{\top})}_{A^{\textrm{sym}}} + \underbrace{\tfrac{1}{2}(A - A^{\top})}_{A^{\textrm{skew}}} ,$$ the invariants of $A^{\textrm{sym}}$ and $A^{\textrm{skew}}$ under congruence are invariants of $A$ under congruence. So the problem of classifying matrices up to congruence is the same as the problem of classifying pairs of matrices, one symmetric and one skew-symmetric, up to congruence by a common matrix, that is, classifying pairs $(A^{\textrm{sym}}, A^{\textrm{skew}})$ up to equivalence under $$(A^{\textrm{sym}}, A^{\textrm{skew}}) \sim (P^{\top} A^{\textrm{sym}} P, P^{\top} A^{\textrm{skew}} P) .$$
Example For complex matrices, the only congruence invariants of $A^{\textrm{sym}}$ and $A^{\textrm{skew}}$ (separately) are their ranks, and these (pairs of) ranks are discrete invariants of a congruence class. In the special case $n = 2$, the possible building blocks are $$\pmatrix{0}, \qquad \pmatrix{1}, \qquad \pmatrix{&1\\-1&}, \qquad \pmatrix{1&-\alpha\\ \alpha&1}, \quad \alpha \in \Bbb C ,$$ which give the canonical forms $$\underset{(0, 0)}{\pmatrix{0&\\&0}}, \quad \underset{(1, 0)}{\pmatrix{1&\\&0}}, \quad \underset{(2, 0)}{\pmatrix{1&\\&1}}, \quad \underset{(0, 2)}{\pmatrix{&1\\-1&}}, \quad \underset{(1, 2)}{\pmatrix{1&-i\\i&1}}, \quad \underset{(2, 2)}{\pmatrix{1&-\alpha\\\alpha&1}}, \,\,\,\, \alpha \neq i .$$ (The parameters $\pm \alpha$ give congruent matrices.) The pairs underset under each form are the invariants $(\operatorname{rank}(A^{\textrm{sym}}) , \operatorname{rank}(A^{\textrm{skew}}))$; the last case is generic. One can show that for a matrix $A$ whose symmetric and skew-symmetric parts have rank $2$, $\alpha^2 = \det A^{\operatorname{sym}} / \det A^{\textrm{skew}}$. The second and third classes are symmetric, the fourth class is skew-symmetric, and the last two classes are neither.
Over $\Bbb R$, most of these classes (or more precisely their intersections with $M(2, \Bbb R)$) decompose into finer real congruence classes; we should expect this, since over $\Bbb R$ the symmetric part alone has signature. Over $\Bbb R$, the real congruence classes consist of the six symmetric classes you already know, the class containing $$\pmatrix{-1&-1\\1&1},$$ and the generic classes $$\pmatrix{1&-\alpha\\\alpha&1}, \qquad \pmatrix{-1&-\beta\\\beta&1}, \quad \beta^2 \neq 1, \qquad \textrm{and} \qquad \pmatrix{-1&-\gamma\\ \gamma&-1}.$$ The matrices corresponding to $\pm \alpha$ are congruent, and likewise for the parameters $\beta$ and $\gamma$. See this note of G.D. Williams for more information about the $2 \times 2$ case over general fields.