If $A,B$ are positive definite, then $\sum a_{ij}b_{ij}\geq 0$

Question

I wish to prove it without using Schur product theorem. Consider the case for $2\times 2$ matrices. I have to show $$a_{11}b_{11}+a_{12}b_{12}+a_{21}b_{21}+a_{22}b_{22}\geq 0$$ Since $B$ is positive definite I can write $b_{11}=b_1b_1$ and $b_{22}=b_2b_2$. Then I have $$a_{11}b_1b_1+a_{12}b_1b_2+a_{21}b_2b_1+a_{22}b_2b_2+a_{12}b_{12}+a_{21}b_{21}-a_{12}b_1b_2-a_{21}b_2b_1$$ and $$a_{11}b_1b_1+a_{12}b_1b_2+a_{21}b_2b_1+a_{22}b_2b_2\geq 0$$ But this doesn't seem to go anywhere. I also have $a_{11}+a_{22}\geq a_{12}+a_{21}$ but that doesn't help me either. Any hint would be appreciated.

Answer 1

Use the formula $S=\sum a_{ij}b_{ij} = \mathrm{tr}(AB)$ to write $S=\mathrm{tr} (UDU' B) =\mathrm{tr} (D U'B U)$, where $D$ is diagonal and $U$ is orthogonal with $A=UDU'$. The diagonal entries of $D$ and of $U'BU$ are positive, so $S>0$.