A matrix with curious property: near-equality of trace and squared Frobenius norm

Amatrixwithcuriouspropertynear-equalityoftraceandsquaredFrobeniusnorm3.gif

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Just a quick question from an observation: is there a reason Tr[A] ~ d/E ?

Nevermind I found it in your comment on the math.SE post

I can show that expected trace is equal to (1- \frac{1}{d})^d …

So for large d, (1- \frac{1}{d})^d \to \frac{1}{e}.

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It appears like there’s a tiny but non-zero difference between the two. P1 and P2 formulas in https://math.stackexchange.com/a/5017119/998

Verification:

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