diagonalization of symmetric bilinear function

[yifli]

According to duality principle, a bilinear function \theta:V\times V \rightarrow R 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 T(i,j)=\theta(\alpha_i,\alpha_j). And this matrix T is diagonalizable, i.e., \theta(\alpha_i,\alpha_i)=0,1,-1.

I don't understand how come \theta(\alpha_i,\alpha_i)=-1

[micromass]

Hi yifli! :smile:

According to duality principle, a bilinear function \theta:V\times V \rightarrow R 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 T(i,j)=\theta(\alpha_i,\alpha_j). And this matrix T is diagonalizable, i.e., \theta(\alpha_i,\alpha_i)=0,1,-1.

I don't understand how come \theta(\alpha_i,\alpha_i)=-1

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:

\theta(x,y)=xy

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

\frac{v}{\sqrt{\theta(v,v)}}

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

However, let's pick

\theta(x,y)=-xy

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

\theta(v,v)=-1

So if we try to normalize this, we get

\frac{v}{\sqrt{\theta(v,v)}}=\frac{v}{\sqrt{-1}}

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.

[Petr Mugver]

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)

[micromass]

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!