How to determine smallest $M>0$ such that $(Au,v)≤M\Vert u \Vert \Vert v \Vert?$

Question

Let $V=\mathbb R^2$ provided with the euclidean inner product and norm. Given a linear, bounded and strong positive operator $$A=\begin{bmatrix} 1 & 1 \\ \frac{1}{2} & 3 \end{bmatrix} $$ is it true that the greatest $\mu>0 $ such that $$(Au,u) ≥ \mu\Vert u\Vert^2$$ is the greatest eigenvalue of A as I assume? Furthermore, how can I determine the smallest $M>0$ such that $$(Au,v)≤M\Vert u \Vert \Vert v \Vert?$$ I have the assumption to somehow involve the symmetric part $\frac{1}{2}(A+A^T)$ to show is as I have seen something similar before. Some help or hints would be much appreciated.

Answer 1

As indicated in the comments below the answer, the $\mu$ in question will be the smallest eigenvalue of $B = (A + A^T)/2$. One explanation is as follows:

• For all $u$, we have $(Au,u) = (Bu,u)$
• By the Rayleigh-Ritz theorem (described briefly in the opening blurb here), the minimum value of $(Bu,u)/\|u\|^2$ is the lowest eigenvalue of $B$.

As for $M$: we begin by noting that the largest singular value of $A$ can be described as $$ \sigma_1(A) = \|A\| = \max_{\|x\| = 1} \|Ax\| $$ With that and the Cauchy-Schwarz inequality, we note that $$ (Au,v) \leq \|Au\| \cdot \|v\| \leq \sigma_1(A)\cdot \|u\| \cdot \|v\| $$ To see that this upper bound is attained (i.e. that there exist $u,v$ such that $(Au,v) = \sigma_1(A)\|u\| \|v\|$), it suffices to take a $u$ such that $\|Au\| = \sigma_1(A)$ and take $v = Au$.