Quadratic forms and sylvester's law of inertia

In summary, the symmetric matrix representing the quadratic form wrt the standard basis for \mathbb{R}^3 is: A =\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}
  • #1
Ted123
446
0
Say I start with a quadratic form:

[itex]x^2 - y^2 - 2z^2 + 2xz - 4yz[/itex].

I complete the square to get:

[itex](x+z)^2 - (y+2z)^2 + z^2[/itex].

(So the rank=3, signature=1)

The symmetric matrix representing the quadratic form wrt the standard basis for [itex]\mathbb{R}^3[/itex] is

[itex]A =\begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ 1 & -2 & -2 \end{bmatrix}[/itex]

I want to find a change of basis matrix [itex]P[/itex] such that

[itex]P^T A P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/itex]

(the 'Sylvester' form - 1s, -1s, 0s on diagonal, 0s elsewhere).

From the complete the square form we can write the quadratic form in the coordinate basis (x+z), (y+2z), z:

[itex]\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}[/itex]

Inverting this gives:

[itex]P = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}[/itex]

(Note that we don't need to scale the columns here)

and we see that we get what we want.

However: what happens if you get a quadratic form that is not of full rank and you want to end up with 0s in the Sylvester form?

e.g. [itex]x^2 + 2xy + 2xz[/itex]

Completing the square gives:

[itex](x+y+z)^2 - (y+z)^2[/itex]

so the rank=2, signature=0.

The symmetric matrix representing the quadratic form wrt the standard basis for [itex]\mathbb{R}^3[/itex] is

[itex]A =\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}[/itex]

We seek a change-of-basis matrix [itex]P[/itex] such that we get

[itex]P^T AP = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/itex]

(the 'Sylvester' form).

Writing the complete the square form in a coordinate basis (x+y+z), (y+z), ... - where does the other basis vector come from?
 
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  • #2
I haven't followed everything through, but in a sylvester form, the number of 1,-1 and 0's are an invariant of the quadratic form and the same as the number of positive, negative and zero eigenvectors in the original matrix.

So in the final case, the only difference is you have an eigenvalue of zero, corresponding to the kernel of the original matrix.
 
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  • #3
as to finding the final basis vector, I haven't tried it yet, but why not try something perpindicular to the other 2...
 
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  • #4
As for another way to get the matrix P, you could find the eigenvalues, then eigenvectors thus giving you the required P matrix

so for the last case you get something like
lambda =-1, (1,-1,-1)
lambda = 0, (0,1,-1)
lambda = 2, (2,1,1)

These vectors define the eigenbasis and can be used to get P and probably work back to the "basis" as you describe it above.

That said the connection between the complete the square basis and the eigenvectors is not totally clear to me, and i would need to work it through
 
  • #5
lanedance said:
As for another way to get the matrix P, you could find the eigenvalues, then eigenvectors thus giving you the required P matrix

so for the last case you get something like
lambda =-1, (1,-1,-1)
lambda = 0, (0,1,-1)
lambda = 2, (2,1,1)

These vectors define the eigenbasis and can be used to get P and probably work back to the "basis" as you describe it above.

That said the connection between the complete the square basis and the eigenvectors is not totally clear to me, and i would need to work it through

But if you put the eigenvectors as columns in P, you get a diagonal matrix with eigenvalues on the diagonal which is not in general in Sylvester form (if you have eigenvalues that are not 0, 1 or -1)
 
  • #6
well say you had an orthonormal matrix of eigenvectors S, that diagonalises A:
[tex] D = S^TDS[/tex]

now say the diagonal entries are the eigenvectors Dii, then define the matrix E, by
[tex] E_{ii} = \frac{1}{\sqrt{|D_{ii}|}} \ \ , \ \ when \ \ D_{ii}\neq0 [/tex]
[tex] E_{ij} = 0 , \ \ otherwise [/tex]

and consider
[tex] M = E^T DE = E^T S^TASE = (SE)^TA(SE)[/tex]

Hence
P = SE
 
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  • #7
So I'm still trying to get my head around the first method.

so start with the quadratic form
[tex] Q_x x^2 - y^2 - 2z^2 + 2xz - 4yz = \textbf{x}^T A \textbf{x} = \textbf{x}^T\begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & -2 \\ 1 & -2 & -2 \end{bmatrix}\textbf{x}
[/tex]

note that quadratic form can be diagonalised, by the transformation that satifisies the completed square, so let [itex] \textbf{u}^T = (u,v,w)[/itex], then
[tex] x^2 - y^2 - 2z^2 + 2xz - 4yz = (x+z)^2 - (y+2z)^2 + z^2 = u^2+v^2+w^2[/tex]

the transformation as you have defined it is given by, each of the row vector is a basis vector
[tex] \textbf{u} = B \textbf{x} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \textbf{x} [/tex]

multiplying by the inverse gives
[tex] \textbf{x} = B^{-1}\textbf{u} = P\textbf{u} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} \textbf{u}
[/tex]

now subtituting into the original quadratic form, to get in terms of the basis u, we get
[tex] Q_x = \textbf{x}^T A \textbf{x} [/tex]
[tex] Q_u = \textbf{u}^T P^TA P\textbf{u} [/tex]
 

FAQ: Quadratic forms and sylvester's law of inertia

1. What is a quadratic form?

A quadratic form is a mathematical expression in the form of ax^2 + bx + c, where a, b, and c are constants and x is a variable. It can also be written in the form of x^T A x, where x is a vector and A is a matrix. Quadratic forms are used in many areas of mathematics and physics, including optimization problems and describing the behavior of systems.

2. What is Sylvester's Law of Inertia?

Sylvester's Law of Inertia, also known as Sylvester's Law of Inertia of Inertia, is a theorem in linear algebra that states that the number of positive, negative, and zero eigenvalues of a symmetric matrix is the same, regardless of the choice of basis. In other words, the number of positive and negative squares in the quadratic form associated with the matrix is independent of the coordinate system used to represent it.

3. How is Sylvester's Law of Inertia related to quadratic forms?

Sylvester's Law of Inertia is directly related to quadratic forms because it provides a powerful tool for analyzing the behavior of quadratic forms. By determining the number of positive, negative, and zero eigenvalues of a matrix, we can determine the type and behavior of the associated quadratic form.

4. What are the applications of quadratic forms and Sylvester's Law of Inertia?

Quadratic forms and Sylvester's Law of Inertia have numerous applications in mathematics and physics. They are used in optimization problems, statistics, mechanics, and quantum mechanics, to name a few. They are also used in machine learning and data analysis to identify patterns and relationships in data.

5. Are there any limitations or exceptions to Sylvester's Law of Inertia?

While Sylvester's Law of Inertia is a powerful and widely applicable theorem, there are some limitations and exceptions. It only applies to symmetric matrices, and it does not provide information about the specific values of the eigenvalues. Additionally, it does not apply to complex matrices. In some cases, it may also be necessary to use other techniques in conjunction with Sylvester's Law of Inertia to fully analyze a quadratic form.

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