The set-up is as follows: I have a complex, Hermitian matrix $H$ with $\mbox{Tr }H=0$, and such that the trace norm $\|H\|_1=1$ (i.e. the sum of the singular values $=1$).
Let me define the functiona $\mbox{TrAbs}$ to ojcemp
am
the sum of the absolute values of the diagonal entries of a matrix (which is clearly dependent on the basis), i.e.
$$ \mbox{TrAbs }H = \sum_{i=1}^n |(H)_{ii}| $$
where $n$ is the dimension of the matrix.
My question is to understand the following: Given $\epsilon$, if I write $H$ in a randomly chosen orthonormal basis and call it $H^\prime$, what is the probability that $\mbox{TrAbs }H^\prime>\epsilon$?
To clarify a couple of things:
- I'm looking for a lower bound on the probability---the exact value is not particularly relevant.
- "randomly" choosing a basis is quite vague. Perhaps this could be done by conjugating with a Haar random unitary, but if there are other ways of choosing a basis randomly that yield a more straightforward bound, I'm sure those would be fine too.
What I know already:
- $\mbox{TrAbs }H^\prime \geq |\mbox{Tr }H|=0$.
- If $H^\prime$ is in its diagonal basis, then clearly $\mbox{TrAbs }H^\prime=\|H\|_1=1$.
- It is then easy to show that $0 \leq \mbox{TrAbs }H^\prime \leq 1$.
I intuitively feel that the probability ought to approach $1$ as the dimension of $H$ tends to $\infty$ (because it seems like there would be so many more bases that give a $\mbox{TrAbs}$ value of $\frac{1}{2}$, say, than would give $0$), but can't show it. So I feel like there should be a way of giving a lower bound that only involves $\epsilon$, and not $n$.
If I can provide any other information about this question, please let me know.
I really appreciate any help you can provide.
I don't see any work. Your business is (for example) to study the simplest case $n=2$.
We assume that $A=\begin{pmatrix}-a&\alpha\\\bar{\alpha}&a\end{pmatrix}$ is a hermitian $2\times 2$ matrix that satisfies $||A||_1=2\sqrt{a^2+|\alpha|^2}=1$, that is $(a,|\alpha|)$ is a point of the circle $(O,1/2)$; we put $a+i|\alpha|=1/2e^{i\theta}$ where $\theta$ follows the uniform law on $[0,\pi]$. Here, $trAbs(A)=2|a|$ and we seek $prob(|a|>\epsilon/2)$. Since $a=1/2\cos(\theta)$,
$prob(|a|>\epsilon/2)=prob(|\cos(\theta)|>\epsilon)=\dfrac{2\arccos(\epsilon)}{\pi}$.
EDIT. I misread the question. In fact, you fix the matrix $H$ and you consider the matrices that are unitarily similar to $H$. Then you may assume that $H=diag((\lambda_i))$ and $prob(trAbs(H')>\epsilon)$ is a function of the $(\lambda_i)$. My friend, it's time to treat the case $n=2$. On the other hand, to study the required prob. (for any $n$), you need to consider a measure (a Haar measure ?) on $SU(n)$. I wish you luck with it.