ksaimouli
  • ksaimouli
orthonormal basis
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ksaimouli
  • ksaimouli
|dw:1449438732751:dw|
amoodarya
  • amoodarya
this is not full rank !
0487308
  • 0487308
(1\/sqrt(3), 1\/sqrt(3), 1\/sqrt(3)) | (0, 0, 0) | (-1\/sqrt(6), -1\/sqrt(6), sqrt(2\/3)) | (0, 0, 0)

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ksaimouli
  • ksaimouli
find orthonormal basis for subspace
ksaimouli
  • ksaimouli
@amoodarya yes, so it does not form a basis 3 dimensional subspace of r^4
0487308
  • 0487308
\[\ (\frac{ 1 }{ \sqrt{3} }, \frac{ 1 }{ \sqrt{3} },\frac{ 1 }{ \sqrt{3} }) | (0,0,0)|(-\frac{ 1 }{ \sqrt{6} }, -\frac{ 1 }{ \sqrt{6} },\sqrt{\frac{ 2 }{ 3 }} |(0,0,0)\]
0487308
  • 0487308
I think that is the orthanormalized version.
0487308
  • 0487308
(0.57735, 0.57735, 0.57735) | (0, 0, 0) | (-0.408248, -0.408248, 0.816497) | (0, 0, 0) There is the orthanormalized basis.
amoodarya
  • amoodarya
which decomposition you are going to do ? lu lq QR Cholesky Decomposition ?
ksaimouli
  • ksaimouli
QR
amoodarya
  • amoodarya
0.707 -0.707 0.000 0.707 0.707 0.000 0.000 0.000 1.000 0.000 0.000 0.000 this is Q matrix 1.414 1.414 1.414 0.000 0.000 0.000 0.000 0.000 1.000 and this is R matrix A=QR
ksaimouli
  • ksaimouli
how did you get that?
amoodarya
  • amoodarya
I can't explain ,how to to do that completely . but I suggest see , your text book .or for a fast result see this https://en.wikipedia.org/wiki/QR_decomposition
ksaimouli
  • ksaimouli
Thanks, I'll. But in general first you used gram-schmidt and then normalized the vectors right?
amoodarya
  • amoodarya
yes . gram-schmidt for 3 or 4 or less than 10 ...is stable to use .
ksaimouli
  • ksaimouli
Okay, thanks I'll try. SECOND topic prove: if A is symmetric and positive de nite, then it is invertible
ksaimouli
  • ksaimouli
(part of the proof) Assuming that A is symmetric and positive de nite, and it is not invertible, then it has a non-trivial null space (why?).
amoodarya
  • amoodarya
I think if A is + definite ,at least rank is 1 ...so has a non-trivial null space . (as honestly as possible :I am not sure ,because many years ago ...I read them)
ksaimouli
  • ksaimouli
No problem, thanks for helping i'll try the above question. So the R is the orthonormal basis right
ksaimouli
  • ksaimouli
but to do a QR decomp. we need to have a lin. indep. vectors right? do i need to find null space of the above vectors then?
amoodarya
  • amoodarya
In practice you can change 1 1 1 1 1 1 to 1 1 1 1 1 1.0001 and do decomposition for this new (almost the same as first one) . and now they are lin-indep

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