Is the converse of Sylvester's inertia law true?

Question

Let $A,B$ be $n\times n$ symmetric matrices.

Assume the positive,negative,0 index of $A$ and $B$ are the same. (That is, they have the same inertia)

Then, are $A,B$ congruent?

What is a counterexample?

Answer 1

Yes, this is true (over the real and complex numbers). If $A$ and $B$ have the same inertia, then they are both congruent to the same diagonal matrix of $1$s, $-1$s, and $0$s. By transitivity, they are congruent to each other.

Answer 2

It depends on the base field you are working with.

For a field like the reals or complex numbers, where every positive element is a square, the answer is yes. There are two changes of basis that make both into the Gram matrix with 1's, 0's and -1's on the diagonal according to index, and by transitivity both matrices are cogredient.

For an ordered field whose positive elements aren't all squares, I think the answer is no. These matrices have matching indices, but aren't cogredient over $\Bbb Q$:

$\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\begin{bmatrix}2&0\\0&1\end{bmatrix}$