Let $v_1 \dots v_n, w_1 \dots w_n \in H$ an inner product space. I am trying (unsuccesfully) to show that $$ \sum_{i,j=1}^n \langle v_i, v_j \rangle \langle w_i, w_j \rangle \geq 0 .$$ Any hints?

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    $\begingroup$ The matrices $\left(\left<v_i,v_j\right>\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ and $\left(\left<w_i,w_j\right>\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ are positive-semidefinite (being Gram matrices). Now there is a fact saying that the Hadamard product of two positive-semidefinite matrices is again positive-semidefinite. Thus, the matrix $\left(\left<v_i,v_j\right>\left<w_i,w_j\right>\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ is positive-semidefinite, and your cliam follows. Is there a simpler proof? $\endgroup$ – darij grinberg Sep 4 '15 at 22:42

The vector space spanned by $v_1,\ldots,v_n,w_1,\ldots,w_n$ is finite-dimensional. Therefore, by a change of basis, we may assume that the inner product space is just $\mathbb C^{2n}$ equipped with the usual inner product. Let $u_i=\bar{w}_i$ for each $i$. Now, using the fact that the trace of a product of matrices is invariant under cyclic permutation of the multiplicands, we have \begin{align} \sum_{i,j=1}^n \langle v_i, v_j \rangle \langle w_i, w_j\rangle &=\sum_{i,j=1}^n v_j^\ast v_i u_i^\ast u_j\\ &=\operatorname{trace} \left(\sum_{i,j=1}^n v_i u_i^\ast u_jv_j^\ast\right)\\ &=\operatorname{trace}\left[\left(\sum_{i=1}^n v_i u_i^\ast\right)\left(\sum_{j=1}^n u_jv_j^\ast\right)\right]. \end{align} As a matrix product of the form $A^\ast A$ (i.e. a Gram matrix) is always positive semidefinite, the assertion follows.

Alternatively, the sum in question is of the form $\mathbf1^\ast\left[(V^\ast V)\circ(W^\ast W)\right]\mathbf1$, where $\mathbf 1$ is the all-one vector, $V$ is the augmented matrix of the $v_i$s and $W$ is the augmented matrix of the $w_i$s. As pointed out by darij grinberg in a comment, the Gram matrices $V^\ast V$ and $W^\ast W$ are positive semidefinite. The Schur product theorem guarantees that the Hadamard product of two PSD matrices is again PSD. Hence the result.


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