Xrfgtjtk

Why is this true? I cant seem to understand why the $det(A) = 1$ for this to hold?

It is not indeed the case. We might have $$det (A)=-1$$ since $$det(A^TJA)=det(J)$$yields to $$det(A^T)det(J)det(A)=det(J)$$and if $J$ is invertible (i.e. the determinant is non-zero) we conclude $$det(A)^2=1$$if $J$ is singuler ($det(J)=0$) we cannot determine $det (A)$

I assume the matrices are over the real numbers.

• In general, $det J$ can be anything.




• If $J$ is invertible, then one can show that $det A = pm 1$ (see for example the answer by MostafaAyaz)




• If $J$ is invertible and antisymmetric with $J= -J^T$, then one can show that $det A =1$. The proof of this fact is not completely elementary as it involves the Pfaffian with $$operatorname{Pf}(A^T J A) = det(A) operatorname{Pf}( J) ,.$$
  As $A^T J A =J$ and $J$ is not singular, we follow $det A=1$.

It is not true in all cases. The determinant is multiplicative, , and $det vphantom{A}^{mathrm tmkern -5mu}A=det A$, so
$$det(vphantom{A}^{mathrm tmkern -5mu}AJA)=(det A)^2det J$$
and if it is equal to $det J$ and $det Jne 0$, you can deduce $det A=pm1$.

We assume $det A >0$

The multiplicative property of determinant implies $$ det A^T JA =(det A)^2 det J =det J$$
You may cancel $det J $ and the result follows. We are also-assuming that $det J ne 0$

It is not true in full generality. For instance, if you are dealing with $1×1$ matrices, then matrix multiplication boils down to ordinary multiplication and the equation would be $Acdot Acdot J = J$, so assuming we are working over the rationals (though this can be generalized), we have $J=0$ or $A=1$ or $A=-1$.
In fact, this example can be generalized by applying the determinant to both sides of the equation and noting that $det (A^T)=det(A)$. This gives $det(A)^2 times det(J) = det(J)$, so if $det(J)$ is not zero, then $det(A)=1$ or $det(A)=-1$.