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Two Gram Schmidt Processes

  1. Nov 24, 2013 #1

    I already know the "usual" Gram-Schmidt Process:

    But, then I discovered another Gram-Schmidt process that gives an orthogonal basis (in Matrices and Linear transformations by Charles Cullen):

    Essentially, it is the same except you do not divide by <uj,uj>. Cullen calls this the Gram-Schmidt process, yet it is different. Does it give a different basis?

    The explanations for the usual Gram-Schmidt process are usually visual while the explanation for this one was algebraic. I cannot make sense of it in any of the visual explanations of the Gram-Schmidt process, yet it makes perfect sense algebraically;

    How are these processes different? Or, if they aren't, why?
    Last edited: Nov 24, 2013
  2. jcsd
  3. Nov 25, 2013 #2


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    That definition does not give an orthogonal basis unless ##\langle u_j,u_j\rangle =1 ## for ##j=1,\ldots k-1##. I am not familiar with the text that you cite, but it is probably the case that there is an additional steps where we normalize the basis vectors. For example we can start with

    $$u'_1 = v_1 , ~~~~ u_1 = \frac{u'_1 }{||u_1'||}.$$

    Then we can construct

    $$ u'_2 = v_2 - \langle v_2,u_1 \rangle u_1$$

    which is orthogonal to ##u_1##, since

    $$ \langle u_1, u'_2 \rangle = \langle u_1,v_2 \rangle - \langle v_2,u_1 \rangle \langle u_1, u_1\rangle =0.$$

    We will have to similarly normalize ##u_2 = u_1'/||u_2'||## if we want to construct ##u_3## with a similar formula that omits the factor of ##\langle u_j,u_j\rangle##.

    The algebraic computation seems clear and also indicates precisely why we need the factor of ##\langle u_j,u_j\rangle## in the formula for the projection operator.
  4. Nov 25, 2013 #3
    I had to reread the text and redo the computations to see that, but yes, it does indeed have to be an orthonormal basis. I just didn't pick up the part that said orthonormal instead of orthogonal.
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