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Suppose $A,B\in \mathbb{R}^{n \times n}$ are non negative definite matrices. We have already know that $A-B$ is also non negative definite. How to show that $\det(A) \geq \det(B)$, if $\det(A)$ means the determinant of matrix $A$?

Zifeng Zhang
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  • Note that askers are expected to provide context for their questions, as is explained here. Please [edit] your question to tell us where you found this question, what you have tried so far, and any other thoughts you might have on the question. – Ben Grossmann Apr 04 '22 at 03:17
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    There's a proof at https://math.stackexchange.com/questions/404965/prove-positive-definite-matrix-and-determinant-inequality also https://math.stackexchange.com/questions/3539312/determinant-inequality-positive-definite-matrices and https://math.stackexchange.com/questions/3515308/if-a-and-b-are-linear-transformations-on-a-finite-dimensional-inner-product – Gerry Myerson Apr 04 '22 at 04:08
  • And probably several others. – Gerry Myerson Apr 04 '22 at 04:14
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    Thank you so much! – Zifeng Zhang Apr 04 '22 at 04:21

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