I'm constructing a weighted norm of the form $||N||_A = \sqrt{tr(N^T A N)}=\sqrt{N_{ij}A_{jk}N_{kl}\delta_{il}}$, where $\{A,N\} \in \mathbb{R}^{3x3}$. In order to keep the argument of the square root positive, I want to be sure that the quantity $tr(N^T A N)\geq 0$, which is to say that $A$ is positive semi-definite with respect to $N$.
For a general matrix $N$, the condition is likely that $A$ must be symmetric positive semi-definite, i.e. no negative eigenvalues.
However, due to the underlying physics of the problem, I am certain that $N$ is symmetric and traceless, and $A$ will be symmetric. I believe this loosens the positive semi-definite condition. For example, I believe that the matrix $$ A = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & -0.1 \end{bmatrix} $$ will produce a positive trace for all symmetric & traceless $N$, despite the negative eigenvalue.
- Is it possible to develop conditions on $A$ that work for this problem for (1a) diagonal $A$ or (1b) symmetric, full $A$?
- If the norm is instead performed with a fourth-order tensor, $||N||_B = \sqrt{N : B : N} = \sqrt{N_{ij}B_{ijkl}N_{kl}}$, can conditions on $B_{ijkl}$ (with major and minor symmetries) be formed?