Surfaces of Quadratic forms

Really tuck with this question:

What type of surface is defined by H = {(x,y,z) | q(x,y,z) = 20 x}

All the definitions of surfaces I know involve terms in x^2. There is a theorem which states that all quadratic forms can be expressed in some basis as the sum of the squares of the basis coordinates, but the method for doing this assumes there is some xy terms or equivalent.

In the derivation of all quadric surfaces being isomorphic to some expression in x^2 there was a line which says: "if a = 0 (the coefficient of x^2) then we can permute the coordinates s.t. a is not equal to 0." But I dont understand how or why this is the case.

I have an exam on wednesday so any help before then would be much appreciated, thanks.

Reply 1

SimonM

18

ax^2+2bxy+2cxz+dy^2+2eyz+fz^2 = [x y z] * M [x y z]^T

where M is the matrix [a b c; b d e; c e f]

Since the spectral theorem tells us we can diagonalize M as QDQ^T, which is equivalent to changing the basis by Q and having an expression just in terms for the quadratic terms

Reply 2

BlueSam3

17

I'll take an example:

q(x,y,z) = 4x^2 - 3y^2 + 2z^2 + yz

. Then the equation defining your surface is:

4x^2 - 20x - 8y^2 + 4z^2 +yz = 0

. Completing the square on the

x

terms gives

(2x-5)^2-8y^2+4z^2 +yz=-25

. Letting

x' = 2x-5

, the LHS of this is now a quadratic form (which I shall call

p

) in

x',y,z

:

p: (x',y,z)\mapsto x'^2-8y^2+4z^2+4yz

. Continuing to normalise this, we obtain

p(x',y,z)=x'^2 + (2z + y)^2 - (3y)^2

. Letting

y' = 3y, z' = 2z+y

, we obtain the normalised form of our quadratic:

q': (x',y',z')\mapsto x'^2 -y'^2 + z'^2

, and our equation becomes

y'^2-x^2-z'^2 = 25

(I've reversed signs for ease of recognition, but haven't bothered to scale away the 25, but obviously, you could do it by just scaling all coordinates by a factor of 5). This, you should be able to recognise as the equation of a hyperboloid of two sheets.

In case you can't recognise these standard quadratic forms, here's a quick list of normalised forms (up to permutations of the variables):

2D:

Ellipse:

x^2+y^2=1

Parabola:

x^2=y

Hyperbola:

x^2-y^2=1

3D:

Ellipsoid:

x^2+y^2+z^2=1

Hyperboloid of one sheet:

x^2+y^2-z^2=1

Hyperboloid of two sheets:

x^2-y^2-z^2=1

Elliptic cone:

x^2+y^2-z^2=0

Elliptic paraboloid:

x^2+y^2=z

Hyperbolic paraboloid:

x^2-y^2=z

Elliptic cylinder:

x^2+y^2=1

Parabolic cylinder:

x^2=y

Hyperbolic cylinder:

x^2-y^2=1

(edited 10 years ago)

Reply 3

puffa

OP

thank you for your help, but I would be able to solve this if it was in the form you have given since I am familiar with this method. My issue however is the equation is q(v) = 20 x, which has no square terms and no terms involving xy, xz, or yz, hence completing the square and other simple coordinate changes don't work. This was a question from last years exam so I assume that it is not simply an error

Reply 4

BlueSam3

17

Original post by puffa

thank you for your help, but I would be able to solve this if it was in the form you have given since I am familiar with this method. My issue however is the equation is q(v) = 20 x, which has no square terms and no terms involving xy, xz, or yz, hence completing the square and other simple coordinate changes don't work. This was a question from last years exam so I assume that it is not simply an error

Look further up the page. Somewhere up there, you'll have

q

defined. Stick it into your equation.

Surfaces of Quadratic forms

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