Matrix/determinant inequality

Question

I have studied the inequality that if $A-B$ is positive semi-definite, then $\det(A)\geq \det(B).$ I was trying to prove the other way around. That if we know that $A$ and $B$ psd and that $\det(A)\geq \det(B)$ then is $A-B$ psd?

I was using eigenvalue method but was unable to complete. However the simulations for random matrices seems to comply.

For $A\succeq B\implies\det(A)\geq\det(B)$, you can use Weyl's inequalities. In particular, you want to show that $\lambda_i(A)\geq\lambda_i(B)$, where $\lambda_i$ are ordered eigenvalues of the symmetric/Hermitian arguments. — Algebraic Pavel, May 03 '15 at 05:41

score 2 · Answer 1 · answered May 01 '15 at 21:22

2

The statement "If $A$ and $B$ are psd and $\det(A) \ge \det(B)$ then $A-B$ is psd" is false.

Consider the matrices $A = \begin{bmatrix}3&0\\0&1\end{bmatrix}$ and $B = \begin{bmatrix}1&0\\0&2\end{bmatrix}$.

answered May 01 '15 at 21:22

JimmyK4542

54,331

Matrix/determinant inequality

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