Hadamard's inequality for Gram determinant

Question

I would like to show that for the Gram determinant $\Gamma(x_1,\ldots,x_n)$ ($(x_1,\ldots,x_n)$ linear independent) we have $$\Gamma(x_1,\ldots,x_n)\leq\prod_{i=1}^{n}||x_i||.$$ Does it have anything to do with the fact that for any squared Matrix $A\in M_{n,n}(R)$ with columns $a_i$ we have $$|\det(A)|\leq \prod_{i=1}^{n}||a_i||?$$ I've found both inequalities under the name Hadamard's inequality, but I can't figure out how one could derive one inequality from the other.

Answer 1

I do not think the first inequality holds. Consider $x_1 = \begin{pmatrix} 2 \\ 0\end{pmatrix}$ and $x_2=\begin{pmatrix}0 \\ 2 \end{pmatrix}$. The Gram matrix will be given by $G = \begin{pmatrix} \langle x_1, x_1 \rangle & \langle x_1,x_2 \rangle \\ \langle x_2, x_1 \rangle & \langle x_2, x_2 \rangle \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}$, hence the Gram determinant is $16$. However, $||x_1|| \times ||x_2||= 2 \times 2 = 4$.

The Gram determinant $\det G$ does satisfy a similar inequality that can be derived by Hadamard's inequality (the second inequality you give). Let $X$ be the matrix whose columns are $x_1, x_2, \dots x_n$. Then the Gram matrix $G$ is given by $G = X^\top X$. Thus, $\det G = (\det X)^2 \leq \left( \prod_{i=1}^n ||x_i|| \right)^2 = \prod_{i=1}^n ||x_i||^2$.