Let S be a symmetric positive n×n matrix and $$B \in M_n(R)$$ a triangularizable matrix with spectrum in $$[0, 1]$$ Prove the inequality: $$tr(BS)\ge tr(B)\det(S)$$
2026-03-26 13:50:43.1774533043
Trace inequality for a symmetric matrix
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This is not true. Consider $$ B=\pmatrix{0&1\\ 0&0},\ S=\pmatrix{\frac12&-\frac14\\ -\frac14&\frac12}. $$ The spectrum of $S$ is $\{\frac14,\frac34\}$ and hence $S$ is positive definite, but $\operatorname{tr}(BS)=-\frac14<0=\operatorname{tr}(B)\det(S)$.