A basic fact in linear algebra is that an N-dimensional vector space can have at most N linearly independent vectors. Is there a better statement of non-orthogonality when there are more vectors than dimensions? This is the best I could derive: Suppose $v_1\dots v_M$ are $M$ normalized vectors in $\mathbb{R}^N$ (that is, $v_i\cdot v_i = 1$ for all $i$), and $M \geq N$. Define $$f = \sum_{i=1}^M\sum_{j=1}^M(v_i\cdot v_j)^2 . $$ Then $f\geq \frac{M^2}{N}$.
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I think you expression is $|V^TV|_F^2$, right? – Thomas Ahle Nov 13 '23 at 07:15
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Maybe look at https://math.stackexchange.com/questions/1013922/constructing-an-epsilon-net-of-l-2-unit-ball – Thomas Ahle Nov 13 '23 at 07:19
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Yes, that's correct, its $|V^T V|_F^2$ – Max Aifer Nov 13 '23 at 08:26