$$A = \left(\begin{array}{cccc} 1&0 \\ 0&-1 \end{array}\right),
B=\left(\begin{array}{cccc} 1&0 \\ 0&2 \end{array}\right), C =
\left(\begin{array}{cccc} 1&0 \\ 0&4 \end{array}\right)$$Are those matrices congruent
- above $\mathbb{R}$?
- above $\mathbb{C}$?
- above $\mathbb{Q}$?
So basically I've evaluated:
$\det (1) = 1, \det \left(\begin{array}{cccc} 1&0 \\ 0&-1 \end{array}\right) = -1$. hence, $A$ is congruent to $\text{Diag}(1,-1)$.
In the same way, $B, C$ are congruent to $\text{Diag}(1,1)$.
Hence, $B$ and $C$ are congruence but, $A$ isn't congruent to $B,C$
This evaluation is above $\mathbb{R}$, but it's also applied to the extended field, $\mathbb{C}$. Isn't it?
Finally, what sohuld I do for $F=\mathbb{Q}$?
Best Answer
Over $\mathbb{R}$ $B$ is congruent to $C$ and as you said above $A$ is not congruent to $B$ and $C$ since the signature of $A$ is $(1,1,0)$ ad the signature of $B,C$ is $(2,0,0)$.
Over $\mathbb{C}$ the situation is quite different. Infact $A$ is equivalent to $B$ and $C$; clearly $B$ is still equivalent to $C$ since real coefficient are also complex coefficient and so the congruence matrix is the same of the case above.
Take $$P= \begin{pmatrix} 1&0\\ 0&i\sqrt{2} \end{pmatrix} $$ so $B=PA\text{}^tP$.
If we call $Q$ the matrix that realize the congruence in the real case we also have $C=QB\text{}^tQ=QPA\text{}^tP\text{}^tQ$.
In the rational case the congruence matrix need to have all coefficients in $\mathbb{Q}$ and you can easily see (using the determinant properties) that there are no matrix that realize the congruence between every couple of matrix above.