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!

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!

 

[blue_raver22]

or \theta(\alpha_i,\alpha_i)=1, as mentioned in the content. Can you please provide more explanation?

Sure, I'd be happy to provide more explanation. The statement that \theta(\alpha_i,\alpha_i)=-1 or \theta(\alpha_i,\alpha_i)=1 is a result of the symmetric property of the bilinear function \theta. This means that for any two vectors \alpha_i and \alpha_j in the vector space V, \theta(\alpha_i,\alpha_j) = \theta(\alpha_j,\alpha_i). In other words, the order of the vectors in the bilinear function does not affect the result.

Now, let's consider the linear mapping T from V to its dual space V*. Since T is a matrix, we can write it in the form of a matrix multiplication: T = \alpha_i^T \cdot \alpha_j. This means that the entry in the i-th row and j-th column of T is given by \theta(\alpha_i,\alpha_j). However, since \theta is symmetric, we know that T(i,j) = T(j,i). This means that the matrix T is symmetric, and therefore can be diagonalized.

To diagonalize a matrix, we need to find a basis for V such that the linear transformation represented by the matrix has a diagonal form. In this case, since T is symmetric, it can be diagonalized by an orthogonal transformation, which means that the basis vectors will be orthogonal to each other.

Now, since T(i,j) = \theta(\alpha_i,\alpha_j), we know that T(i,i) = \theta(\alpha_i,\alpha_i). And since we are looking for a diagonal matrix, we want all the off-diagonal entries to be zero. This means that T(i,i) = 0 for all i. However, since T is a symmetric matrix, we can choose the basis vectors such that T(i,i) = 1 or T(i,i) = -1. This is because we can scale the basis vectors without affecting the diagonalization process.

Therefore, \theta(\alpha_i,\alpha_i) = 0, 1, or -1, depending on the choice of basis vectors. This is why the content mentions that \theta(\alpha_i,\alpha_i) can have these values. I hope this helps clarify the concept of diagonalization of symmetric bilinear functions.