Timeline for An upper estimate for $|\det(A+B)|$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 29, 2018 at 0:24 | vote | accept | Piotr Hajlasz | ||
| Dec 24, 2018 at 12:51 | comment | added | Suvrit | The bound $|\det(A+B)|=\det|A+B| \le \prod_i (a_i+b_{n-i+1})$, where $a_i$ and $b_i$ are the singular values of $A, B$, respectively (sorted in the usual decreasing order) easily holds. Moreover, it follows from the following even more elegant inequality: $|\det(A+B)|^2 \le \det(|A|+|B|)\det(|A^*|+|B^*|)$, where $|A|$ denotes the matrix absolute value ($|A|=(A^*A)^{1/2}$). | |
| Dec 24, 2018 at 7:28 | answer | added | C.F.G | timeline score: 6 | |
| Dec 24, 2018 at 1:55 | answer | added | Aleksei Kulikov | timeline score: 4 | |
| Dec 24, 2018 at 1:11 | comment | added | Aleksei Kulikov | Are you interested in some generalizations of this inequality or best possible $C(n)$? I have reasonably short proof that $C(n) = \frac{2^{n-1}}{\sqrt{n}^n}$ (that is, $A = B = I$ is optimal), but if you want something else could you please specify it somehow. | |
| Dec 23, 2018 at 23:35 | history | asked | Piotr Hajlasz | CC BY-SA 4.0 |