Clearing entries column-by-column, as opposed to in increasing triangles. I tried to illustrate the the order, but managed to mistype it: should have read 21, 31, 41, ..., 32, 42, ..., 43, ... WFL
On 5/18/16, Warren D Smith <warren.wds@gmail.com> wrote:
to FWL, perhaps your working by columns not rows is since you transposed the matrices versus my convention? In any case, doesn't really matter.
On 5/18/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< one may lead a horse to water ... >> --- and if one leaves the dumb animal there long enough, it may eventually get the idea!
Warren's algorithm doesn't quite work as presented: I did try it, by the way. However, working column-by-column instead it succeeds: eliminate in order 21, 31, 31, ..., 32, 42, ..., 43, ... and an even isometry is indeed magically reduced a product of (general rather than adjacent) Givens rotations.
The Wikipedia page paragraph "Stable calculation" discusses tinkering with the sign of the rotation angle in order to ensure "continuity". Does anyone understand what this is about?
WFL
On 5/17/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
See https://en.wikipedia.org/wiki/Givens_rotation (one may take a horse to water ... )
WFL
On 5/17/16, Warren D Smith <warren.wds@gmail.com> wrote:
LUNNON replying to WDS: 7 48 259 136...
Rotation #3 destroys the result of #1 ?!
As I pointed out previously, Givens etc. reduction is _iterative_ --- each operation destroys previous entries, though leaving those smaller than they had been earlier. WFL
--WDS. You're totally wrong. First of all Givens transforms need NOT be done "iteratively" meaning an infinite process. A finite number of them can be used, and are used, in the schemes I described, to exactly zero all the entries of the lower triangle of any desired starting orthogonal nXn matrix, one by one, without ever destroying any previous zero. To accomplish that it is key to perform the zeroings in a feasible well-chosen order.
Second, rotation #3 does NOT destroy the zero-result caused by rotation 1 because when 3 is used, entries labeled 1 and 2 have already been zeroed by the two prior Givens's. Therefore this rotation, affecting the bottom two rows only, does not alter that fact the entries labeled 1 and 2 already both contain 0s, because of the amazing mathematical facts that 0+0=0 and x*0=0 and 0-0=0.
Next?
As should have been obvious, each GIvens, because of your demand it only affect rows i and i+1 of the nXn matrix, only affects rows i and i+1 of the nXn matrix, for different i each time of course.
You appear (trying to read your mind) to have the idea that Givens's can be used iteratively from both the left and the right (conjugately) to zero the strict lower and upper triangle of a symmetric matrix, eventually yielding its eigenvalues on diagonal and rest 0s. That is true. That iterative method was first suggested by Jacobi. It generally requires an infinite number of Givens's to complete, although a finite number suffice to accomplish the zeroing to any particular number of decimal places accuracy. However, that all is completely irrelevant. I am working on an orthogonal, not a symmetric, nXn matrix, and multiplying by Givens's only from the left not both left & right, and zeroing entries one at a time, not an iterative approximate-zeroing.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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