Majorizing the even distribution
Anders B Madsen 26 Feb 2024 20:32
The eigenvalue vector $\lambda^{\psi}$ of $\operatorname{\bf tr}_B \ket{\psi}\bra{\psi}$ becomes
(1)\begin{align} \lambda^{\psi} = (\frac1m, \frac1m,\ldots,\frac1m)\,. \end{align}
So all we have to show is that if
(2)\begin{align} \sum_{i=1}^m x_i = 1\,, \quad x_i \geq 0\,,\quad x_i \geq x_{i+1} \end{align}
then $(x_1,\ldots,x_m)$ majorizes $(\frac1m,\ldots,\frac1m)$.
For $k \leq m$, it holds for the partial sum $S_k$ that
(3)\begin{align} S_k = \sum_{i=1}^k x_i \geq kx_k\,, \end{align}
since $x_k$ is the smallest of the terms. Likewise,
(4)\begin{align} 1-S_k = \sum_{i=k+1}^m x_i \leq (m-k)x_k \leq (m-k) \frac{S_k}k\,, \end{align}
since $x_k$ is larger than any of the terms, and where also the first inequality was used.
This leads to $1 \leq \frac{m}k S_k$, or
(5)\begin{align} S_k \geq \frac{k}m\,, \end{align}
which shows that $(x_1,\ldots,x_m)$ majorizes $(\frac1m,\ldots,\frac1m)$.