# A matrix inequality involving the Hilbert-Schmidt norm

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.

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Very naive question: What happens if you differentiate $Q$ with respect to $v$ and solve for critical points? In principle, you should be able to identify the minimum value of $Q(v)$ in terms of $a$ this way. – Deane Yang Dec 10 '12 at 16:16
Not naive at all. I tried with the analytic approach but it did not seem to clarify the problem, indeed $Q(v)$ is a fourth order polynomial in $v$. Before plunging in the computations, I was hoping to find some more synthetic approach using matrix inequalities – Piero D'Ancona Dec 10 '12 at 16:35
Using coordinates that make $a$ diagonal probably makes this simpler. – Noam D. Elkies Dec 10 '12 at 16:55

## 1 Answer

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 $a=\lambda I_3$.

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That was quick :) Thanks! – Piero D'Ancona Dec 10 '12 at 17:03