- #1

Ted123

- 446

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[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?