This question comes from a problem in PDEs on which I'm currently working. Let $a$ be a $3\times 3$ matrix, real symmetric and positive definite. Denote with $\|a\|^2 \_ 2=\sum a\_{ij}^2$ the square of the Hilbert-Schmidt norm and consider the quantity
$$ Q(v)= 2\|a\|^2 \_2 + [trace(a)-3(av,v)]^2
 -6[2 |av|^2 -(av,v)^2] $$
where $v$ is an arbitrary unit vector. If $a=I$ is the identity, the quantity $Q$ is identically zero. 


QUESTION: are there other matrices $a$ such that $Q(v)\ge0$ for all unit vectors $v$, or is this condition equivalent to $a=I$?


At least, it would be helpful if the matrix wizards around here could suggest ways to handle the HS norm, which is unfamiliar to me, and estimates relating trace, HS norm, operator norm and the numerical range of a matrix, which could possibly be of use here.



A matrix inequality involving the Hilbert-Schmidt norm
Suppose $Q$ is such a form. Write that the mean value of $Q$ over the unit sphere is non-negative. You obtain
$$-\frac12\sum\_ia\_{ii}^2+\frac12\sum\_{i < j}a\_{ii}a\_{jj}-9\sum\_{i < j}a\_{ij}^2\ge0.$$
This implies that $a=\lambda I\_3$.


