Gram determinant How to prove that
$$\sqrt{\Gamma(\vec{a},\vec{b},\vec{c})}=|(\vec{a},\vec{b},\vec{c})|,$$
where $\Gamma(\vec{a},\vec{b},\vec{c})= \left | \begin{array} {ccc} \vec{a} \cdot \vec{a} & \vec{b} \cdot \vec{a} & \vec{c} \cdot \vec{a} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} & \vec{c} \cdot \vec{b} \\
\vec{a} \cdot \vec{c} & \vec{b} \cdot \vec{c} & \vec{c} \cdot \vec{c}
\end{array} \right | $.
Thanks!
 A: Let the matrix $X$ have rows $\vec{a},\vec{b},\vec c$.
Then $XX^T$ is the matrix within the vertical bars in your definition of the gram determinant.
Then you have $\Gamma(\vec{a},\vec{b},\vec{c})=\det(XX^T)=\det(X)\det(X^T)=\dots$
Can you take it from here?

If in fact you do want to work with complex vectors, then I need to alter the above slightly. To be consistent with the order of terms in the dot products, you would have ot use column vectors instead of row vectors, and the Hermitian conjugate instead of the transpose. You'd also need the fact that the determinant of $X^H$ is the complex conjugate of the determinant of $X$, and multiplying them would give you the square modulus of the determinant of $X$.
A: Presumably $\vec{a}$ and the like are column vectors. Your $\Gamma$ is just the matrix product $A^\ast A$, where $A=(\vec{a},\vec{b},\vec{c})$ and $A^\ast$ denotes the conjugate transpose of $A$ (or simply the transpose of $A$ if $\vec{a},\vec{b},\vec{c}$ are real). Using the properties that $\det(XY)=\det(X)\det(Y)$ and $\det(X^\ast)=\overline{\det(X)}$, we see that $\det(\Gamma)=\overline{\det(A)}\det(A)=|\det(A)|^2$ and hence the correct statement should be:
$$\sqrt{\det(\Gamma)}=|\det(A)|,$$
where the pair of strokes $|\cdot|$ here denotes the modulus of a complex number (or absolute value of a real number). Without taking the modulus, your statement is wrong. A counterexample is given by $\vec{a}=(-1,0,0)^\top,\ \vec{b}=(0,-1,0)^\top$ and $\vec{c}=(0,0,-1)^\top$.
