Timeline for Continuous change of basis (and on the definition of determinant)
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2012 at 8:27 | comment | added | Gabriel Nivasch | Let me add for the record that there is a simpler way of doing the continuous transformation in the same spirit as the suggestion of Robert Israel. We wish to transform $(u_1,...,u_n)$ into $(e_1,...,e_{n-1},\pm e_n)$. We do Gauss-Jordan elimination continuously, with the restriction that a row cannot be multiplied by a negative number (because the multiplier, which starts at 1 and changes continuously, would have to go through $0$). We get $(\pm e_1,...,\pm e_n)$. Then we eliminate the minus signs in pairs with Givens rotations, as Israel suggested. We might be left with one minus at the end. | |
| Nov 12, 2012 at 22:38 | vote | accept | Gabriel Nivasch | ||
| Nov 12, 2012 at 18:12 | comment | added | Robert Israel | If $R$ is a rotation matrix in ${\mathbb R}^n$, we can multiply it on the left by Givens matrices $G(1,2,\theta_2), G(1,3,\theta_3), \ldots, G(1,n,\theta_n)$ successively making the $(2,1), (3,1), \ldots, (n,1)$ entries $0$. Then the $(1,1)$ entry will be $\pm 1$ and we can make it $+1$ by adding $\pi$ to $\theta_n$ if necessary; the $(1,2), \ldots, (1,n)$ entries will be $0$. Thus now we have a block matrix $\pmatrix{1 & 0\cr 0 & S\cr}$ where $S$ is an $(n-1) \times (n-1)$ rotation matrix. Proceed by induction. | |
| Nov 11, 2012 at 20:59 | comment | added | Gabriel Nivasch | Robert: Do you mean that any rotation in $\mathbb R^n$ can be decomposed into Givens rotations? I see such a statement in the Wikipedia for dimension 3. Is it true in general? | |
| Nov 11, 2012 at 19:16 | history | answered | Robert Israel | CC BY-SA 3.0 |