I have a following question.

Assume that there are two positive definite matrices $B_1$ and $B_2$ such that det$(B_1) \le$ det$(B_2)$ where det$(.)$ denotes the determinant of a matrix. Suppose $A=bb^T$ where all the entries of $b$ are non-zero (thus all diagonal entries of $A$ are strictly greater than zero). I want to show that (if at all it is true) det$(A \circ B_1) \le$ det$(A \circ B_2)$ if and only if det$(B_1) \le$ det$(B_2)$. Is this true? If yes where can I find the proof ($B_1$ and $B_2$ are not related by an order). If not true, is there a counter example? Thanks.


1 Answer 1


This is true. Let $b^t=(b_1,\ldots,b_n)$ and define $D$ as the diagonal matrix with diagonal $b_1,\ldots,b_n$.

For any matrix $C$ of order $n$ notice that $bb^t\circ C=DCD$.

Thus, $\det(bb^t\circ B_i)=\det(DB_iD)=\det(D)^2\det(B_i)$ for $i=1,2$.

Since $\det(D)^2>0$ then $\det(B_1)\leq\det(B_2)$ iff $\det(bb^t\circ B_1)\leq \det(bb^t\circ B_2)$.

  • 1
    $\begingroup$ @Sanand you are welcome. $\endgroup$
    – Daniel
    Sep 21, 2016 at 16:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .