Consider the matrix $A=\begin{bmatrix}0&1\\
1&0
\end{bmatrix}$ over the field $\Bbb Q$ of rationals.Which of the following matrices are congruent to $A$ over $\Bbb Q$?
$(1).$ $A_1=\begin{bmatrix}2&0\\
0&-2
\end{bmatrix}$
$(2).$ $A_2=\begin{bmatrix}2&0\\
0&2
\end{bmatrix}$
$(3).$ $A_3=\begin{bmatrix}1&0\\
0&-1
\end{bmatrix}$
$(4).$ $A_4=\begin{bmatrix}3&4\\
4&5
\end{bmatrix}$
We know that,
Two matrices $A$ and $B$ of size $n\times n$ are congruent over any field $\mathbb{F}$ if there exists $P\in GL(n,\mathbb{F})$ s.t. $A=P^{t}BP$.
Now,
$A=P^{t}BP\implies \frac{det(A)}{det(B)}={(det(P))}^2$.
Since, $det(P)\in\mathbb{F}\implies\pm\sqrt{\frac{det(A)}{det(B)}}\in\mathbb{F}$.
Hence in view of the above discussion $A_2$ is not congruent to $A$(as $\pm\sqrt{\frac{det(A)}{det(A_1)}}\notin\mathbb{Q}$).
Now for other options how should i proceeed?
Is there any necessary and sufficient condition for congruency of two matrices over field of rationals(Like Sylvester's criteria in case of reals)?
If two matrices are congruent over $\mathbb{Q}$ then are their invariants(rank,signature,index) same?
Best Answer
$$ \left( \begin{array}{rr} \frac{3}{2} & \frac{ 1 }{ 2 } \\ \frac{1}{2} & \frac{ 3 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rr} 1 & 0 \\ 0 & -1 \\ \end{array} \right) \left( \begin{array}{rr} \frac{3}{2} & \frac{ 1 }{ 2 } \\ \frac{1}{2} & \frac{ 3 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rr} 2 & 0 \\ 0 & - 2 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$\left( \begin{array}{rr} 1 & 1 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rr} 0 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{rr} 1 & - \frac{ 1 }{ 2 } \\ 1 & \frac{ 1 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rr} 2 & 0 \\ 0 & - \frac{ 1 }{ 2 } \\ \end{array} \right) $$
$$ \left( \begin{array}{rr} \frac{3}{4} & \frac{ 1 }{ 2 } \\ \frac{1}{4} & \frac{ 3 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rr} 2 & 0 \\ 0 & -\frac{1}{2} \\ \end{array} \right) \left( \begin{array}{rr} \frac{3}{4} & \frac{ 1 }{ 4 } \\ \frac{1}{2} & \frac{ 3 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rr} 1 & 0 \\ 0 & - 1 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$\left( \begin{array}{rr} 1 & 0 \\ - \frac{ 4 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rr} 3 & 4 \\ 4 & 5 \\ \end{array} \right) \left( \begin{array}{rr} 1 & - \frac{ 4 }{ 3 } \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 3 & 0 \\ 0 & - \frac{ 1 }{ 3 } \\ \end{array} \right) $$ $$ $$
$$ \left( \begin{array}{rr} \frac{7}{12} & \frac{ 1 }{ 4 } \\ \frac{1}{12} & \frac{ 7 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rr} 3 & 0 \\ 0 & -\frac{1}{3} \\ \end{array} \right) \left( \begin{array}{rr} \frac{7}{12} & \frac{ 1 }{ 12 } \\ \frac{1}{4} & \frac{ 7 }{ 4 } \\ \end{array} \right) = \left( \begin{array}{rr} 1 & 0 \\ 0 & - 1 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ \left( \begin{array}{rr} \frac{n+1}{2n} & \frac{ n-1 }{ 2 } \\ \frac{n-1}{2n} & \frac{ n+1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rr} n & 0 \\ 0 & -\frac{1}{n} \\ \end{array} \right) \left( \begin{array}{rr} \frac{n+1}{2n} & \frac{ n-1 }{ 2n } \\ \frac{n-1}{2} & \frac{ n+1 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rr} 1 & 0 \\ 0 & - 1 \\ \end{array} \right) $$
Note: I was thinking of $n$ as an integer, but this matrix product works fine for any rational nonzero $n.$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$