Suppose we have positive-definite matrices $A$, $B$, if $A>B>0$, can we always find a positive real number $\alpha$, $0<\alpha < 1$ such that $ \alpha A \geq B $? If it has, then what is it?



If $A>B>0$, can we always find a positive real number $\alpha$, $0<\alpha < 1$ such that $\alpha A \geq B $?
It seems the following.


Let $A$ and $B$ be $n\times n$ matrices and $S\subset\Bbb R^n$ be the unit sphere. Then positive definiteness of the matrix $A$ is equivalent to 


$$m(A)=\min\{(Ax,x):x\in S\}>0.$$


Similarly, we have 


$$m(A-B)=\min\{(A-B)x,x):x\in S\}>0.$$


Put 


$$M(B)=\max\{(Bx,x):x\in S\}>0.$$


Let $\alpha>\frac{M(B)}{m(A-B)+M(B)}<1$ be any number and $x\in S$ be any vector. Then 


$$(\alpha Ax,x)-(Bx,x)=$$ $$\alpha((A-B)x,x)-(1-\alpha)(Bx,x)\ge$$ $$\alpha m(A-B)-(1-\alpha)M(B)=$$ $$\alpha(m(A-B)+M(B))-M(B)>$$ $$\frac{M(B)}{m(A-B)+M(B)} (m(A-B)+M(B))-M(B)>0.$$ 


Then 


$$m(\alpha A-B)= \min\{(\alpha A-B)x,x):x\in S\}>0.$$


So the matrix $\alpha A−B$ is positive-definite, that is $\alpha A>B$


