If $A,B \in M_n(\mathbb{R})$, how can I prove that $\det\left(A^TA+B^TB\right)\ge0$? Any hint would be highly appreciated.
2026-03-27 02:35:47.1774578947
How to prove that $\det(A^TA+B^TB) \geq 0$ for $A,B $ real matrices?
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You first show that the sum of positive semi-definite is positive semi-definite, and then that the determinant of positive semi-definite is non-negative.