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Suppose $A$ is positive-definite, symmetric and real. Show that for $u,v\in\mathbb{R}^{n}$ that $$2|Au\cdot v|\leq\rho^{\frac{1}{2}}(u\cdot Au+|v|^{2}),$$ where $\rho$ is the maximal eigenvalue of $A$, e.g. $\rho:=\text{max}(\rho_{1},\ldots,\rho_{n}).$ I've tried fiddling around with different things, but haven't arrived at a concise proof, and I need the result for some other things that I am doing right now, so I figured I'd see if anyone on here had a proof in mind.

Edit (1) (Response to Gary)

At first try, this leads to something similar to what I already obtained by using the identity $||A||_{2}=\sqrt{\rho(AA^{t})}=\sqrt{\rho(A^{2})}=\rho$ (since $A=A^{T}$). In details, $$2|Au\cdot v|\leq2|Au||v|\leq2||A||_{2}|u||v|\leq2\rho|u||v|\leq\rho^{\frac{1}{2}}(|u|^{2}+|v|^{2}).$$ Similarly, following Gary's suggestion, if $A$ is diagonalized and we're in the orthonormal eigenbasis, we have $$2|Au\cdot v|=2\left|\sum\limits_{j=1}^{n}\rho_{j}u_{j}v_{j}\right|\leq2\rho(u\cdot v)\leq2\rho|u||v|\leq\rho^{\frac{1}{2}}(|u|^{2}+|v|^{2}),$$ which is the same estimate as above.

In any case, the proof is reduced to showing $$|u|^{2}\leq u^{T}Au=\sum\limits_{j=1}^{n}u_{j}\rho_{j}u_{j}=\sum\limits_{j=1}^{n}\rho_{j}u_{j}^{2}\leq\rho|u|^{2}.$$

But this inequality can hold if and only if $\rho\geq1$. Is there a theorem which concludes positive definite, real and symmetric matrices necessarily have an eigenvalue greater than or equal to $1$? I doubt this, so either I've made a mistake or the problem that I am working on left out an assumption on $A$...thoughts? Is there a sharper estimate than just the Cauchy inequality used above?

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${\bf R}^n$ will have an orthonormal basis consisting of eigenvectors of $A$. Maybe if you write $u$ and $v$ as linear combinations of these eigenvectors, something will come out. –  Gerry Myerson Feb 28 '13 at 0:39
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I don't think it is the case that$\left|u\right|^{2}\leq u^{T}Au$. Just take $A=\frac{1}{2}I$ where $I$ is the identity matrix. –  Han Altae-Tran Feb 28 '13 at 1:38
    
You're right. There must be a sharper estimate available, or the inequality is false unless $\rho\geq1.$ (Obviously from what has already been shown it is true if this condition holds.) –  Taylor Martin Feb 28 '13 at 2:12

1 Answer 1

Let $B$ be the square root of $A$, then, by the Cauchy Schwartz inequality, for every $u$, $v$ from $\Bbb{R}^n$, we have $$ 2\vert\langle Bu,Bv\rangle\vert\leq 2\Vert Bu\Vert\cdot\Vert Bv\Vert\leq \Vert B\Vert \left(2\Vert Bu\Vert\cdot\Vert v\Vert\right) \leq \Vert B\Vert \left(\Vert Bu\Vert^2+\Vert v\Vert^2\right), $$ but $\langle Bu,Bv\rangle=\langle Au,v\rangle$, $\Vert Bu\Vert^2=\langle Au,u\rangle$, and $\Vert B\Vert=\rho^{1/2}$. This proves the desired inequality.$~\square$

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