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 $? If it has, then what is it?
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$\begingroup$ $A>B$ means $A-B$ is positive-definite matrix. $\endgroup$– wayneCommented Jun 7, 2015 at 3:07
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$\begingroup$ Naturally. Thank you. Can't you prove a little sharper result: $\ \alpha\cdot A\ >\ B\ $ ? $\endgroup$– Włodzimierz HolsztyńskiCommented Jun 7, 2015 at 3:08
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$\begingroup$ If you mean my proposition is always true? $\endgroup$– wayneCommented Jun 7, 2015 at 3:10
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2$\begingroup$ The real part the relevant expression related to $\ A - B\ $ is greater than certain $\ \epsilon > 0\ $ when evaluated on the unit sphere. If you change $\ A-B\ $ a little, i.e. if you change $\ A\ $ a little then it'll be fine. $\endgroup$– Włodzimierz HolsztyńskiCommented Jun 7, 2015 at 3:29
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$\begingroup$ Thanks for you suggestion, after following your suggestion, I arrange it as follows: $A-B>\epsilon I$ which is equivalent to $A-\epsilon I> B$, also $(I-\epsilon A^{-1}) A> B$, but this still can't show $\alpha A>B$ where $0< \alpha <1$. $\endgroup$– wayneCommented Jun 7, 2015 at 3:45
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2 Answers
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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 $\alpha A>B$
answered Jun 7, 2015 at 5:17
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$\begingroup$ Thanks for your efforts, I have 2 questions regarding your answer: 1) why let $A$ and $B$ be $n×n$ matrices and $S\subset R^n$ be the unit sphere? 2) $(\alpha Ax, x)-(Bx,x)>0$ may not imply $m(\alpha A-B)>0$. $\endgroup$– wayneCommented Jun 7, 2015 at 7:49
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$\begingroup$ The general definition considers the complex field, not real. $\endgroup$ Commented Jun 7, 2015 at 8:14
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$\begingroup$ Can $\alpha =\frac{M(B)}{M(B)+m(A-B)}$ such that $\alpha A\geq B$? $\endgroup$– wayneCommented Jun 7, 2015 at 12:44
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$\begingroup$ 1) I just fixed $n$. $S$ represents all non-zero vectors of the space $\Bbb R^n$. I considered only a real case, but I hope that the proof for the complex case is similar. $\endgroup$ Commented Jun 8, 2015 at 5:25
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1$\begingroup$ 2) The sphere $S$ is compact, so a continuous positive-valued function on $S$ attains its (positive) infimum (that is why I wrote $”\min”$ instead of $”\inf”$.) $\endgroup$ Commented Jun 8, 2015 at 5:25
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The @Alex Ravsky's answer is good, however, I still want to share my answer:
Proof: To prove $\alpha A\geq B$ where $0<\alpha<1$, we introduce an extra parameter $\lambda$, $0<\lambda<1$, $\alpha=1-\lambda$, such that $(1-\lambda)A\geq B$, which is equivalent to find a $\lambda$, $0<\lambda<1$, such that $(1-\lambda)A\geq B$.
Arrange it, we get, $\lambda A\leq A-B$, to make it hold, we shall find $\lambda$ such that $\lambda A \leq \lambda_{\min}(A-B)I$ holds, which is equivalent to $\lambda I \leq \lambda_{\min}(A-B) A^{-1}$. Further, let $\lambda = \frac{\lambda_{\min}(A-B)}{\lambda_{\max}(A)}$, the above equations are automatically satisfied.
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$\begingroup$ I expect that, in fact, these $\lambda_{m\dots}$’s are the respective (minimal of maximal) eigenvalues of the matrices. $\endgroup$ Commented Jun 8, 2015 at 5:30
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1$\begingroup$ 1) I expect the following. Let $C$ be a positive definite symmetric $n\times n$ real-value matrix. Then $$\min\{(Cx,x):x\in S\}=\lambda_{min}(C)$$ and $$\max\{(Cx,x):x\in S\}=\lambda_{max}(C),$$ where $\lambda_{min}$ and $\lambda_{max}$ are, respectively, minimal and maximal eigenvalues of the matrix $C$. $\endgroup$ Commented Jun 8, 2015 at 10:09
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1$\begingroup$ Thanks for your comments, it's clear now. Regarding your reply 1): I don't know $\lambda_{\min}$ or $\lambda_{\max}$ can be interpreted in this form. 3): I tried and arrived at $\frac{\lambda_{\max}(B)}{\lambda_{\max}(B)+\lambda_{\min}(A-B)} - \frac{\lambda_{\max}(A)-\lambda_{\min}(A-B)}{\lambda_{\max}(A)}=\frac{\lambda_{\min}(A-B)(\lambda_{\min}(A-B)-\lambda_{\max}(A)+\lambda_{\max}(B))}{\lambda_{\max}(A)\lambda_{\max}(B)+\lambda_{\max}(A)\lambda_{\min}(A-B)}$. Since it's hard to decide whether $(\lambda_{\min}(A-B)-\lambda_{\max}(A)+\lambda_{\max}(B))>0$ or not, I stop here. $\endgroup$– wayneCommented Jun 8, 2015 at 14:46
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1$\begingroup$ 3) It seems the following. Yes, I arrived to this inequality too. From here we can proceed as follows. Let $x\in S$ be an arbitrary vector. Then $$(Ax, x)-(Bx, x)=((A-B)x , x)\ge\lambda_{min}(A-B)$$ and $$(Ax , x)\le\lambda_{max}(A).$$ So $$(Bx, x)\le \lambda_{max}(A)- \lambda_{min}(A-B).$$ Since the vector $x$ was chosen arbitrarily, we may claim that $$\lambda_{max} (B)\le \lambda_{max}(A)- \lambda_{min}(A-B).$$ $\endgroup$ Commented Jun 8, 2015 at 20:12
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1$\begingroup$ Thanks for your efforts, everything is clear now. It seems that your bound is smaller than mine. $\endgroup$– wayneCommented Jun 9, 2015 at 1:00