I have a following question.
Assume that there are two positive definite matrices $B_1$ and $B_2$ such that $\lambda_{min}(B_1) \le \lambda_{min}(B_2)$ where $\lambda_{min}(.)$ denotes the minimum eigenvalue 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 $\lambda_{min}(A \circ B_1) \le \lambda_{min}(A \circ B_2)$ if and only if $\lambda_{min}(B_1) \le \lambda_{min}(B_2)$. Is this true? If yes where can I find the proof. ($B_1$ and $B_2$ are not related by an order) Thanks.
This is not true. Let $B_1=\pmatrix{3 & 0\\0 & 1 }$, $B_2=\pmatrix{2 & 0\\0 & 2 }$ and $b^t=(\frac{1}{2}, 1)$.
Notice that $bb^t=\pmatrix{\frac{1}{4} & \frac{1}{2}\\\frac{1}{2} & 1 }$, $bb^t\circ B_1=\pmatrix{\frac{3}{4} & 0\\0 & 1 }$ and $bb^t\circ B_2=\pmatrix{\frac{1}{2} & 0\\0 & 2 }$.