Timeline for A direct proof of a property of symmetric 2x2-determinants
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2017 at 10:47 | history | edited | Dima Pasechnik | CC BY-SA 3.0 |
small formatting fix
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| Jun 22, 2017 at 8:41 | vote | accept | Dima Pasechnik | ||
| Jun 22, 2017 at 8:39 | comment | added | Dima Pasechnik | If $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}=\sum_k\begin{pmatrix}p_k^2&p_kq_k\\p_kq_k&q_k^2\end{pmatrix}$ then $\det M$ is an s.o.s., as $\det M=\sum_i\sum_j(p_iq_j-p_jq_i)^2.$ In particular, if $M$ is a psd matrix of forms, it's not clear whether the above decomposition of $M$ as a sum of matrices of forms exists, or not. | |
| Jun 21, 2017 at 14:20 | comment | added | darij grinberg | Well, the statement is that if $A$ and $B$ are positive semidefinite symmetric matrices, then so is $A+B$. But how do you encode positive semidefiniteness of a matrix $C$ as a statement about s.o.s.? As "the matrix is a sum of matrices of the form $D^T D$" or as "all principal minors of the matrix are sums of squares"? The former makes the claim really easy. The latter seems more complicated -- you are asking to use s.o.s. decompositions of unknown provenance; I suspect there are no general results about this, but I don't know a good way to generate counterexamples either. | |
| Jun 21, 2017 at 11:24 | comment | added | Dima Pasechnik | Thanks. AM-GM is useful. Regarding sums of squares (s.o.s.) decompositions, as Ivan noted it's a positive semidefinite matrix, and so the corresponding quadratic form over $\mathbb{R}$ is a s.o.s. Now, replace $\mathbb{R}$ with a ring of real $m$-variate polynomials; if $m=1$ then our determinant will be an s.o.s. of polynomials, as any nonnegative univariate polynomial is an s.o.s. How about $m>1$? | |
| Jun 21, 2017 at 10:31 | history | answered | darij grinberg | CC BY-SA 3.0 |