Consider $3 \times 3$ matrix $A$. The sum of squares of all entries of $A$ is not greater than $1$. What is the maximum value of $\det(A)$?
2026-04-04 23:26:28.1775345188
Maximal determinant of a matrix with $\| A \|_F = 1$
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Hint: In terms of the singular values $\sigma_1,\sigma_2,\sigma_3$ of $A$, we have $$ |\det(A)| = \sigma_1\sigma_2\sigma_3, \quad \|A\|_F^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2. $$