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Gram-Schmidt Orthogonalization

Module by: Steven Cox

Summary: (Blank Abstract)

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Suppose that MM is an mm-dimensional subspace with basis x1…xm x1 … xm We transform this into an orthonormal basis q1…qm q1 … qm for MM via

  1. Set y1=x1 y1 x1 and q1=y1∥y1∥ q1 y1 y1
  2. y2=x2 y2 x2 minus the projection of x2x2 onto the line spanned by q1q1. That is y2=x2−q1q1Tq1-1q1Tx2=x2−q1q1Tx2 y2 x2 q1 q1 q1 q1 x2 x2 q1 q1 x2 Set q2=y2∥y2∥ q2 y2 y2 and Q2=q1q2 Q2 q1 q2 .
  3. y3=x3 y3 x3 minus the projection of x3x3 onto the plane spanned by q1q1 and q2q2. That is y3=x3−Q2Q2TQ2-1Q2Tx3=x3−q1q1Tx3 y3 x3 Q2 Q2 Q2 Q2 x3 x3 q1 q1 x3 q2 q2 x3 Set q3=y3∥y3∥ q3 y3 y3 and Q3=q1q2q3 Q3 q1 q2 q3 . Continue in this fashion through step (mm).
  • (mm) ym=xm ym xm minus its projection onto the subspace spanned by the columns of Q m − 1 Q m − 1 . That is ym=xm− Q m − 1 Q m − 1 T Q m − 1 -1 Q m − 1 Txm =xm−∑j=1m−1qjqjTxm ym xm Q m − 1 Q m − 1 Q m − 1 Q m − 1 xm xm j 1 m 1 qj qj xm
Set qm=ym∥ym∥ qm ym ym To take a simple example, let us orthogonalize the following basis for ℝ3 3 , x1=100 x2=110 x3=111 x1 1 0 0 x2 1 1 0 x3 1 1 1
  1. q1=y1=x1 q1 y1 x1 .
  2. y2=x2−q1q1Tx2=010T y2 x2 q1 q1 x2 0 1 0 and so, q2=y2 q2 y2 .
  3. y3=x3−q1q1Tx3=001T y3 x3 q1 q1 x3 q2 q2 x3 0 0 1 and so, q3=y3 q3 y3 .
We have arrived at q1=100 q2=010 q3=001 q1 1 0 0 q2 0 1 0 q3 0 0 1 . Once the idea is grasped the actual calculations are best left to a machine. Matlab accomplishes this via the orth command. Its implementation is a bit more sophisticated than a blind run through our steps (1) through (mm). As a result, there is no guarantee that it will return the same basis. For example 


          >>X=[1 1 1;0 1 1 ;0 0 1];

          >>Q=orth(X)

          Q=

           0.7370  -0.5910  0.3280

           0.5910   0.3280 -0.7370

           0.3280   0.7370  0.5910

This ambiguity does not bother us, for one orthogonal basis is as good as another. Let us put this into practice, via (10.8).

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