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yifli

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

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- Thread starter yifli
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In summary, according to the 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 be represented as a matrix T such that T(i,j)=\theta(\alpha_i,\alpha_j). This matrix T is diagonalizable, meaning it can be represented by a diagonal matrix with 0, 1, or -1 on the diagonal. However, this -1 arises because the real numbers are not algebraically closed, and in the complex numbers, T can be represented with 1's on the diagonal. This is equivalent to picking an orthogonal base for theta, which can be normalized to an orthonormal base. However,

- #1

yifli

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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!

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.

yifli said:

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.

Last edited:

- #3

Petr Mugver

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Petr Mugver said:

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

- #5

blue_raver22

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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.

The diagonalization of symmetric bilinear function is a mathematical process that transforms a symmetric bilinear function into a diagonal form. It involves finding a basis for the vector space on which the function is defined, such that the function can be expressed as a sum of simpler, diagonal functions.

Diagonalization of symmetric bilinear function is important because it simplifies the representation of the function, making it easier to analyze and understand. It also allows for efficient computation of the function, which is useful in various scientific and engineering applications.

The process of diagonalization involves finding the eigenvalues and eigenvectors of the symmetric bilinear function. The eigenvectors form a basis for the vector space, and the eigenvalues correspond to the diagonal elements of the diagonalized form of the function.

Diagonalization of symmetric bilinear function has various applications in the fields of linear algebra, physics, and engineering. It is used in the analysis of systems with quadratic forms, solving systems of linear differential equations, and in diagonalizing matrices for efficient computation.

Yes, all symmetric bilinear functions can be diagonalized. This is because symmetric bilinear functions are defined on a vector space with a finite number of dimensions, and there exists a basis for this vector space that diagonalizes the function. However, the diagonal form may have complex eigenvalues and eigenvectors in some cases.

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