Timeline for Determinant and restriction of scalar
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6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2020 at 16:01 | comment | added | Espace' etale | Very good, wonderful! Infact the blocks will in general be members of an embedding of $E$ inside the ring of matrices over $E$ of size $[E:F]$, right? | |
| Mar 31, 2020 at 13:18 | comment | added | darij grinberg | @Espace'etale: You want to represent it as an $n\times n$ block matrix made out of commuting $2 \times 2$-blocks, not as a $2\times 2$ block matrix made out of commuting $n \times n$-blocks. Each $2 \times 2$-block will have the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ for two real numbers $a$ and $b$; but all such blocks commute. | |
| Mar 31, 2020 at 13:12 | comment | added | Espace' etale | @darijgrinberg it is not true that these blocks commute. For instance if $A+iB$ is a complex $n\times n$ matrix, then the matrix representing it over $\mathbb{R}$ is the $2\times 2$ block matrix with blocks $A ,-B ,B ,A$, but $A,B$ don't commute. I don't see how theorem 1 implies the original question. | |
| Sep 6, 2018 at 11:03 | comment | added | Fred Rohrer | This is a special case of Bourbaki, Algèbre III.9.4 Proposition 6. | |
| Sep 6, 2018 at 4:26 | comment | added | darij grinberg | As with many things, this follows from Theorem 1 in John R. Silvester, Determinants of Block Matrices, The Mathematical Gazette, Vol. 84, No. 501 (Nov., 2000), pp. 460--467. In fact, the matrix that represents $T$ as an $F$-linear operator on $V$ is a block matrix, whose blocks are elements of $F$ written as square matrices over $E$; clearly these blocks commute. Separability of $E/F$ is unnecessary. | |
| Sep 6, 2018 at 3:50 | history | asked | Windi | CC BY-SA 4.0 |