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Matrix decomposition with NullSpace and QRDecomposit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64159] Matrix decomposition with NullSpace and QRDecomposit
  • From: Tobias Burnus <burnus at gmx.de>
  • Date: Thu, 2 Feb 2006 19:09:12 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello,

I have a Hermitian matrix of the form

( A    B )
( B'   C ) =: M

(B' = Transpose[B]), of which I want to bring B (5=D718) into this form
{D, O} =

 d11 d12 d13 d14 d15 0 0 0 0 0 0 0 0 0 0 0 0 0
 d21 d22 d23 d24 d25 0 0 0 0 0 0 0 0 0 0 0 0 0
 d31 d32 d33 d34 d35 0 0 0 0 0 0 0 0 0 0 0 0 0
 d41 d42 d43 d44 d45 0 0 0 0 0 0 0 0 0 0 0 0 0
 d51 d52 d53 d54 d55 0 0 0 0 0 0 0 0 0 0 0 0 0

where O should be zero. For a simple case I succeeded using
  Q = QRDecomposition[B]
  L = NullSpace[Transpose[B]],  (see below)
but it fails for a more complicated case - there not all (O)ij are zero.
(See http://www.physik.fu-berlin.de/~tburnus/tmp/MatrixDecomposition.nb )

This is with Mathematica 5.2.
It works if I set pds1=pds2=...=pds6 and pdp1=...pdp6.

Any ideas?

Besides, does anyone know a quick way to get D Hermitian, currently it
has triangular form.

Tobias

Excerpt from the notebook
--------
HC1 = (.... the matrix ...)
HC1C := HC1[[Range[1, 5], Range[6, 23]]]

Q = QRDecomposition[
        Transpose[HC1C]][[1]]
L = NullSpace[HC1C]
T = IdentityMatrix[Length[HC1]];

T[[Range[6, 6 + Length[Q] - 1],
      Range[6, Length[HC1]]]] = Q;
\!\(\(T\[LeftDoubleBracket]Range[6 + Length[Q], Length[HC1]],
        Range[6, Length[HC1]]\[RightDoubleBracket]\  = \
Table[L\[LeftDoubleBracket]i\[RightDoubleBracket]\/\@\(\(Conjugate[L]\)\
\[LeftDoubleBracket]i\[RightDoubleBracket] . L\[LeftDoubleBracket]i\
\[RightDoubleBracket]\), {i, 1, Length[L]}];\)\)

T=Assuming[
      pd\[Sigma]1<0&&pd\[Pi]1>0&&pd\[Sigma]2<0&&pd\[Pi]2>0&&pd\[Sigma]3<0&&
        pd\[Pi]3>0&&pd\[Sigma]4<0&&pd\[Pi]4>0&&pd\[Sigma]5<0&&pd\[Pi]5>0&&
        pd\[Sigma]6<0&&pd\[Pi]6>0,FullSimplify[T]];

myHC1 = T.HC1.Transpose[T] // FullSimplify


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Cell[BoxData[
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Cell[BoxData[
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Cell[BoxData[
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Cell[BoxData[
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          ColumnSpacings->1,
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Cell[CellGroupData[{

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                    pd\[Sigma]3\^2 + 
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pd\[Sigma]6\^2)\)\)\)\ \@\(\((pd\[Sigma]1\^2 + pd\[Sigma]2\^2 + \
pd\[Sigma]3\^2 + pd\[Sigma]4\^2)\)\ \((\((pd\[Sigma]3\^2 + pd\[Sigma]4\^2)\)\ \
\((pd\[Sigma]5\^2 + pd\[Sigma]6\^2)\) + pd\[Sigma]1\^2\ \((pd\[Sigma]3\^2 + \
pd\[Sigma]4\^2 + pd\[Sigma]5\^2 + pd\[Sigma]6\^2)\) + pd\[Sigma]2\^2\ \((pd\
\[Sigma]3\^2 + pd\[Sigma]4\^2 + pd\[Sigma]5\^2 + pd\[Sigma]6\^2)\))\)\)\)\/\@\
\(\((pd\[Sigma]1\^2 + pd\[Sigma]2\^2 + pd\[Sigma]3\^2 + pd\[Sigma]4\^2)\)\ \
\((\((pd\[Sigma]3\^2 + pd\[Sigma]4\^2)\)\ \((pd\[Sigma]5\^2 + pd\[Sigma]6\^2)\
\) + pd\[Sigma]1\^2\ \((pd\[Sigma]3\^2 + pd\[Sigma]4\^2 + pd\[Sigma]5\^2 + pd\
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