Diagonalization of symmetric bilinear function

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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]
 

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  • #2
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Hi yifli! :smile:

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]

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:

[tex]\theta(x,y)=xy[/tex]

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

[tex]\frac{v}{\sqrt{\theta(v,v)}}[/tex]

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

However, let's pick

[tex]\theta(x,y)=-xy[/tex]

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

[tex]\theta(v,v)=-1[/tex]

So if we try to normalize this, we get

[tex]\frac{v}{\sqrt{\theta(v,v)}}=\frac{v}{\sqrt{-1}}[/tex]

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.
 
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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)
 
  • #4
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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)

Indeed, I've edited my post. Sorry yifli!
 

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