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Eigenvalues of Gramian Matrix

  1. Jan 24, 2015 #1
    if x is a column vector, then a matrix G = x*xT is a Gramian Matrix.
    When I tried calculating the matrix G and its eigenvalues for cases when x = [x1 x2]' and [x1 x2 x3]'
    by actually working out the algebra, it turned out (if I didn't do any mistakes) that the eigen values are all zeros except one which is equal to (x12+x22 OR x12 + x22 + x32) depending upon the case.
    Is this a standard result for a Gramian Matrix to have a single non-zero eigenvalue? If, yes, is there a simpler proof?

    Thank you.
     
  2. jcsd
  3. Jan 24, 2015 #2
    Yes that is a standard and pretty simple fact. How did you compute the eigenvalues? Did you compute characteristic polynomials and then find their roots? If so, then yes, there is a much simpler proof.
     
  4. Jan 24, 2015 #3
    Yes, I solved the roots of characteristic equation, and it was a nasty business for even a 3x3 matrix. :D Would love to know the simpler method.
     
  5. Jan 24, 2015 #4
    Let ##\lambda \ne 0## be an eigenvalue, and ##\mathbf v\ne\mathbf 0## be he corresponding eigenvector. That means ##G\mathbf v =\lambda\mathbf v##. But $$G \mathbf v = \mathbf x (\mathbf x^T \mathbf v)$$ and ##(\mathbf x^T \mathbf v)## is a scalar. Therefore $$(\mathbf x^T \mathbf v) \mathbf x = \lambda \mathbf v,$$ so the eigenvector ##\mathbf v## must be a non-zero multiple of ##\mathbf x##. Substituting ##a\mathbf x## (where ##a\ne0## is a scalar) in the above equation, we get that indeed ## a\mathbf x ## is an eigenvector corresponding to ##\lambda= \mathbf x^T\mathbf x##.
     
  6. Jan 24, 2015 #5
    I cannot see why the bold part should be true?
    But, doing the underlined part, i.e. substitution v=ax, I can see that it gives a solution, is that from this you infered that the bold part should hold?

    Thank you for your help.
     
  7. Jan 25, 2015 #6
    No, the "bold" part is true independently of the "underlined" part, they both prove different parts of the statement.

    For the "bold" part: we know that ##(\mathbf x^T\mathbf v)## is a number, let call it ##\beta##. Then the equation is rewritten as ##\beta\mathbf x = \lambda\mathbf v##, and solving it for ##\mathbf v ## gives us ##\mathbf v = (\beta/\lambda) \mathbf x##.

    Now, the constant ##\beta## depends on the unknown ##\mathbf v##, and we do not know what ##\lambda## is, so we cannot say from here that ##\mathbf v = (\beta/\lambda) \mathbf x= a\mathbf x## is an eigenvector. But what we can say is that if ##\mathbf v## is an eigenvector corresponding to a non-zero eigenvalue ##\lambda##, then it must be a non-zero multiple of ##\mathbf x##.

    Substituting then ##\mathbf v=a\mathbf x## we get that it is indeed an eigenvector and find ##\lambda##. So the "underlined" part give you that ##\mathbf v=a\mathbf x## is an eigenvector, and gives the corresponding eigenvalue. The "bold"" part shown that there are no other eigenvectors corresponding to a non-zero eigenvalue.
     
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