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Diagonalization of symmetric bilinear function

  1. Jun 24, 2011 #1
    According to duality principle, a bilinear function [itex]\theta:V\times V \rightarrow R[/itex] is equivalent to a linear mapping from V to its dual space V*, which can in turn be represented as a matrix T such that [itex]T(i,j)=\theta(\alpha_i,\alpha_j)[/itex]. And this matrix T is diagonalizable, i.e., [itex]\theta(\alpha_i,\alpha_i)=0,1,-1[/itex].

    I don't understand how come [itex]\theta(\alpha_i,\alpha_i)=-1[/itex]
  2. jcsd
  3. Jun 24, 2011 #2
    Hi yifli! :smile:

    That -1 arises because the real numbers are not algebraically closed. In the complex numbers, we have that T is diagonalizable with 0 or 1 on the diagonal.

    Basically, saying that theta is diagonalizable is equivalent to picking an orthogonal base for theta. Let's pick an example:


    This is a bilinear form and the following base is orthogonal: v=(5,0), w=(0,10). However, we can normalize this by doing:


    This yields the base (1,0), (0,1). So we have a diagonalizable matrix with 1's on the diagonal.

    However, let's pick


    this is a bilinear form. An orthogonal base for this is again v=(1,0), w=(0,1). However, for this we have


    So if we try to normalize this, we get


    but this cannot be in the real numbers. It is possible in the complex number, however, and this yields the orthonormal base (-i,0),(0,-i). The norms for this basis are 1, thus we get the matrix with 1's on the diagonal.
    Last edited: Jun 25, 2011
  4. Jun 25, 2011 #3
    Uhm, are you guys confusing the terms "diagonalizable" and "orthonormalizable", or is it me, the one confused? When you diagonalize a matrix, don't you multiply on the left and right with a matrix and it's inverse? Instead, in orthonormalization, don't you multiply on the left and right with a matrix and it's transposed? (that's exactly what happens in Micromass's last example)
  5. Jun 25, 2011 #4
    Indeed, I've edited my post. Sorry yifli!
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