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

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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)