Timeline for Is the trace distance upper bounded by the Euclidean distance?
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| Aug 1, 2023 at 18:01 | comment | added | Quantum Mechanic | @gIS well... that's a better answer than mine! | |
| Aug 1, 2023 at 17:51 | comment | added | glS♦ | You can also strengthen the inequality you cited replacing $d$ with the rank of $A$, which you can prove observing your argument also holds if instead of $I$ you use the projection on the support of $A$ (at least it works for positive semidefinite operators; the argument might need some adjustment for the more general case, I'm not sure). These inequalities are a standard tool discussing t-designs, see eg en.wikipedia.org/wiki/Welch_bounds | |
| Aug 1, 2023 at 17:46 | comment | added | glS♦ | I think you can say something quite a bit stronger: you always have the opposite inequality, $\|A\|_1\ge \|A\|_2$. You see it easily from the fact that $\|A\|_1$ is the sum of the singular values of $A$, while $\|A\|_2^2$ is the sum of the squares of the singular values. Thus $\|A\|_1^2 \ge \|A\|_2^2$, as the square of a sum is always larger than the sum of squares, which is equivalent to $\|A\|_1 \ge \|A\|_2$. | |
| Aug 1, 2023 at 14:17 | history | answered | Quantum Mechanic | CC BY-SA 4.0 |