For positive definite matrices $A$ and $B$ of order $n$, if $A \geq B$ then $\det(A+B) \geq \det(A) + n \det(B)$?

Question

What's the shortest way to answer the following question?

Prove that if $A$ and $B$ are positive definite matrices of order $n$ and $A\ge B$, then $\det(A+B) \ge \det(A) + n\det(B)$.

All I can do is a volume argument about $\det(A+B) \geq \det(A) + \det(B)$, but the $n$ in there gives me problems.

Answer 1

Hint prove it for $B=I_n$ ($tr(I_n)=n$)

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Perhaps the hint was not enough so I will try to complete with an answer

• First case $B = I_n$, let $\lambda_1, \ldots, \lambda_n$ be the eigen values of $A$, then the eigen values of $A+I_n$ are $\lambda_1 + 1, \ldots, \lambda_n+1$ so $$\det(A+I) = \prod_{i=1}^n \left(1+ \lambda_i\right) \ge \prod_{i=1}^n \lambda_i + \sum_{i=1}^n\lambda_i \ge \det(A) + \text{tr}(A) \ge \det(A) + \text{tr}(I_n) = \det(A) + n\det(I_n)$$

• If $B$ is not invertible the inequality is trivial. Otherwise, $B^{-\frac12}A B^{-\frac12}\ge I_n$ applying the above result we have $$\det(B^{-\frac12}A B^{-\frac12} + I_n) \ge \det(B^{-\frac12}A B^{-\frac12}) + n$$ now you just have to multiply by the $\det(B)$