# Very general inner product determinant inequality

Let $V$ be a vector space and $\langle \cdot, \cdot\rangle$ be an inner product on $V$. Prove that for any positive integer $n$ and any $x_1,\dots,x_n \in V$

$$\det \left[ \begin{array}{cccc} \langle x_1, x_1 \rangle & \langle x_2, x_1 \rangle &\dots & \langle x_n, x_1 \rangle\\ \langle x_1, x_2 \rangle & \langle x_2, x_2 \rangle & \dots & \langle x_n, x_2 \rangle\\ \vdots & \vdots & \ddots & \vdots \\ \langle x_1, x_n \rangle & \langle x_2, x_n \rangle & \dots & \langle x_n, x_n \rangle \end{array} \right]\geq 0\,.$$

The case of $n=1$ is trivial, it follows from the inner product's defining property. The case of $n=2$ is true due to the Cauchy-Schwarz inequality.

A less general formula came up for $n=3$ during physics research, where there are physical reasons to expect that this inequality ought to hold. I cannot find a counterexample for any vector space dimension or matrix dimension. Is this fundamental inequality a well-known theorem?

• The Gramian matrix is positive semidefinite. Hence, its determinant is nonnegative. – Rodrigo de Azevedo Aug 25 '16 at 19:13
• If you're familiar with differential forms, it might be worth noting that the inner product $\langle \cdot,\cdot \rangle$ on $V$ canonically induces an inner product on each exterior power $\wedge^k V$ of $V$ by $$\langle x_1 \wedge \cdots \wedge x_k, y_1 \wedge \cdots \wedge y_k \rangle = \det(\langle x_i,y_j\rangle);$$ from this perspective, what you're checking is that $$\langle x_1 \wedge \cdots \wedge x_n,x_1\wedge\cdots\wedge x_n\rangle \geq 0,$$ as one would want for an inner product. – Branimir Ćaćić Aug 26 '16 at 3:22

It is a well-known formula (look up Gram matrix). Let $W={\rm Span }\{x_1,\ldots,x_n\}$ be of dimension $k\leq n$ and pick $(e_i)_{1\leq i\leq k}$ an orthonormal base for $W$. Then you may write:

$$\langle x_i,x_k \rangle = \sum_j\langle x_i, e_j\rangle \langle e_j, x_k\rangle$$ or in terms of matrices (with obvious notation): $$X = M^T M$$ The rank of $M$ whence of $X$ is not greater than $k$ so if $k<n$ the vectors in $X$ must be linearly dependent and $\det(X)=0$. If $k=n$ then $M$ has rank $n$ (show this) and $\det(X)= \det(M^T)\det(M)= (\det(M))^2 > 0$. In particular, $\det(X)=0$ iff the $x_i$'s are linearly dependent.

• This works for the case where the $n=\dim(V)$, but how do you prove the more case of $n<\dim(V)$? In that case $M$ is not a square matrix. – bkocsis Aug 26 '16 at 5:41
• Well spotted. It suffices to look at a subspace including the $x_i$'s. I have added that. This also makes it easy to show the last part about equality. – H. H. Rugh Aug 26 '16 at 5:54
• Thanks, but what do you mean by a subspace including the span of $x_i$ exactly? Can it be any subspace which includes this set? Then the determinant of the right hand side may depend on how many dimensional subspace we chose (i.e. it may be zero or nonzero), while $\det(X)$ is given to be a given number, right? Can you fix this? Also let's allow for $x_i$ to be vectors over the complex field. – bkocsis Aug 26 '16 at 11:01