$A^TJA = J$ $ \rightarrow \det(A) = 1$

Question

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

Answer 1

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 $\det (A)$

Answer 2

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

Answer 3

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 $\det J \ne 0$

Answer 4

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 $\det A=1$.

Answer 5

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 $A\cdot A\cdot J = J$, 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 $\det(A)=-1$.