5
votes
Accepted
Inequality for hermitian matrices
No. For example, let
\begin{align*}
p_1 = \begin{bmatrix}
1 & 0 \\ 0 & 0
\end{bmatrix}, \ \ p_2 = \begin{bmatrix}
0 & 0 \\ 0 & 1
\end{bmatrix}, \ \ S_1 = \begin{bmatrix}
1 & 1 \\ 1 ...
- 6,351
3
votes
Inequality for commuting hermitian operators
Yes. Since $S_1$, $S_2$ commute, there exists an orthonormal basis in which they are both diagonal: $S_1={\rm diag}\,(a_1,\ldots,a_n)$ and $S_2={\rm diag}\,(b_1,\ldots,b_n)$, $a_i\geqslant c_1,b_i\...
- 109k
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