Wikipedia:Reference desk/Archives/Mathematics/2023 November 3

= November 3 =

Limiting proportions of this matrix "metric"
The real-valued "metric" $$d$$ on the set of $$n \times n$$ matrices whose values are all $$0$$ or $$1$$ defined by $$d(A, B) = |\det(A - B)|$$ is, as the air quotes suggest, not actually a metric, as there are matrices one can find that break the triangle inequality with it. However, in my probabilistic attempts to determine the proportion of triplets $$(A, B, C)$$ such that $$d(A, C) < d(A, B) + d(B, C)$$, $$d(A, C) > d(A, B) + d(B, C)$$, and $$d(A, C) = d(A, B) + d(B, C)$$, it appears that the first proportion approaches a value around $$70\%$$, the second proportion tends to $$0$$, and the third proportion naturally tends to $$30\%$$. This leads me to two questions:

1. Is there actually a limit to these proportions and can it be expressed concisely?

2. Are there infinitely many matrix triples $$A, B, C$$ such that $$d(A, C) > d(A, B) + d(B, C)$$, even the proportion of such triples tends to $$0$$?

GalacticShoe (talk) 15:45, 3 November 2023 (UTC)


 * For 2, my suggestion would be to let H be a Hadamard matrix with determinant n^(n/2), with all 1s in the first row. Let A, C be 0,1 matrices with H=A-C. Let B be the zero matrix. Then d(A,C)=n^(n/2), d(B,C)=0 and d(A,B) is exponentially smaller than d(A,C), using the idea in this MO post and the fact that Hadamard matrices maximise the determinant among [-1,1] matrices. —Kusma (talk) 16:20, 3 November 2023 (UTC)
 * This is an excellent idea and it settles question 2 in the affirmative, though for my own future remembrance and clarity, I would like to note specifically that the reason why $$d(B, C) = |\det(B - C)| = |\det(-C)| = |\det(C)| = 0$$ is because $$H$$ having all $$1$$s in the first row means that $$C$$ must have all $$0$$s in the first row. Thanks! GalacticShoe (talk) 19:28, 3 November 2023 (UTC)