For Hermitian matrices $X, Y$, I write $X\ge Y\ge 0$ to mean $X-Y$ and $Y$ are positive semidefinite.

In Lemma 2.5 of [Linear Algebra Appl. 452 (2014) 1-6] I proved that if $X + Y\ge W + Z$, $X\ge W\ge Y\ge 0$ and $X\ge Z \ge Y\ge 0$, then $$\det X+\det Y\ge \det W+\det Z.$$

I was thinking of relaxing the condition a little bit.

Is the same inequality true under only the condition $X + Y\ge W + Z$, $X\ge Y\ge 0$, $X\ge W\ge 0$, $X\ge Z\ge 0$?

  • 4
    $\begingroup$ WLOG, $X=I$. After that, WLOG, $Y$ is diagonal. After that, forget about determinants for $Z,W$, just look at the products of diagonal entries. Am I missing anything? $\endgroup$
    – fedja
    Jul 1, 2014 at 3:56
  • $\begingroup$ @fedja: It is right to assume $X=I$ and $Y=D$ to be diagonal. Oh, it seems that I have an answer. $\endgroup$
    – M. Lin
    Jul 1, 2014 at 19:10

1 Answer 1


The answer is yes. I thank fedja for discussion.

WLOG, assume $X=I$ (we may do so by pre-post multiplying both sides by $\det X^{-1/2}$, with a standard continuity argument).

After this, we may further assume $Y=D$ to be diagonal (by unitary similarity).

So the question is equivalent to showing if $I+D\ge W+Z$ and $I\ge D, W, Z\ge 0$, then $1+\det D\ge \det W+\det Z$.

Note that $I+D\ge W+Z$ implies $I+D\ge diag(W)+diag(W)$, where $diag(\cdot)$ means the diagonal part.

By a simple induction on the size of the matrix, one gets $1+\det D\ge \det (diag(W))+\det (diag(Z))$. The conclusion follows by applying the Hadamard inequality, $\det (diag(W))\ge \det W$.

  • $\begingroup$ Could you give more details of the induction step? Thanks in advance. $\endgroup$
    – Tadashi
    Jul 13, 2014 at 20:27

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