Principal axis transformation/Real quadrics/2/Section

Definition

A quadratic polynomial ${}F\in K[X_{1},\ldots ,X_{n}]$ over a field ${}K$ is a polynomial of degree ${}2$; that is, an expression of the form

${}F=\sum _{i\leq j}a_{ij}X_{i}X_{j}+\sum _{i=1}^{n}b_{i}X_{i}+c\,,$

where

${}a_{ij},b_{i},c\in K$.

Example

For a quadratic polynomial ${}aX^{2}+bX+c$ in one variable ${}X$, with ${}a,b,c\in K$ and ${}a\neq 0$, the zeroes can be found by completing the square. That is, we write (suppose that the characteristic of the fields is not ${}2$)

${}aX^{2}+bX+c=a{\left(X^{2}+{\frac {b}{a}}X+{\frac {c}{a}}\right)}=a{\left({\left(X+{\frac {b}{2a}}\right)}^{2}-{\frac {b^{2}}{4a^{2}}}+{\frac {c}{a}}\right)}\,.$

This equals ${}0$ if and only if

${}X=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}-{\frac {b}{2a}}\,,$

and if the square root

${}{\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}={\frac {1}{2a}}{\sqrt {b^{2}-4ac}}\,$

exists in the field. Depending on this, there are no, one or two solutions.

We develop a relation between quadratic polynomials and bilinear forms.

Definition

For a bilinear form ${}\left\langle -,-\right\rangle$ on a ${}K$-vector space ${}V$, the mapping

$V\longrightarrow K,v\longmapsto \left\langle v,v\right\rangle$

is called the

corresponding quadratic form.

Given a basis ${}v_{1},\ldots ,v_{n}$, a bilinear form is described by its Gram matrix

${}G={\left(g_{ij}\right)}_{ij}\,;$

the corresponding quadratic form ${}V\rightarrow K$ is described by the quadratic polynomial

${}\left(X_{1},\,\ldots ,\,X_{n}\right)G{\begin{pmatrix}X_{1}\\\vdots \\X_{n}\end{pmatrix}}=\sum _{1\leq i,j\leq n}g_{ij}X_{i}X_{j}=\sum _{i}g_{ii}X_{i}^{2}+\sum _{i<j}{\left(g_{ij}+g_{ji}\right)}X_{i}X_{j}\,.$

(${}X_{i}$ is the ${}i$-th projection, the corresponding dual basis). In the symmetric case, this is

$\sum _{i}g_{ii}X_{i}^{2}+\sum _{i<j}2g_{ij}X_{i}X_{j}.$

For every pure-quadratic polynomial in ${}n$ variables, we can form in this way a symmetric Gram matrix. The theory of real-symmetric bilinear forms makes it possible to get rid, by a suitable coordinate transformation (a base change) of the mixed terms.

Example

We make a list of real quadratic polynomials in the two variables ${}X$ and ${}Y$, together with their corresponding vanishing sets; we restrict to coefficients from ${}0,1,-1$. If only one variable ${}X$ occurs, then we have essentially the following three possibilities.

• ${}X^{2}\,$ ${}\,$ ${}\,$ the vanishing set is a "doubled line“.
• ${}X^{2}-1\,$ ${}\,$ ${}\,$ this means
${}X=\pm 1$, the vanishing set consists in two parallel lines.
• ${}X^{2}+1\,$ ${}\,$ ${}\,$ the vanishing set is leer.

In these cases (where the second variable ${}Y$ does not occur explicitly), the vanishing set is simply the product set of a zero-dimensional vanishing set (finitely many points) and of a line.

Now we consider polynomials where both variables occur.

• ${}Y^{2}-X\,$ ${}\,$ ${}\,$ the vanishing set is a parabola.
• ${}Y^{2}-X^{2}\,$ ${}\,$ ${}\,$ this means
${}(Y-X)(Y+X)=0$, the vanishing set consists in two lines crossing each other.
• ${}Y^{2}+X^{2}\,$ ${}\,$ ${}\,$ the only solution is the point ${}(0,0)$, the vanishing set is just a single point.
• ${}Y^{2}-X^{2}-1\,$ ${}\,$ ${}\,$ this means
${}(Y-X)(Y+X)=1$, the vanishing set is a hyperbola.
• ${}Y^{2}+X^{2}-1\,$ ${}\,$ ${}\,$ the vanishing sets is the unit circle.
• ${}Y^{2}+X^{2}+1\,$ ${}\,$ ${}\,$ again, this is empty.

The polynomial ${}XY-1$ does not appear directly in this list, because, in the variables ${}X=U+V$ and ${}Y=U-V$, we can write

${}U^{2}-V^{2}=1\,.$

In this form it is in the lists. The following theorem tells us that, up to scaling of the individual variables, the list is complete.

A paraboloid.

A hyperbolic paraboloid, also called a saddle surface.

An ellipsoid. Its surface is a quadric.

A double cone.

A one-sheeted hyperboloid.

A two-sheeted hyperboloid.

Theorem

Every real quadratic polynomial

${}F=\sum _{i\leq j}a_{ij}X_{i}X_{j}+\sum _{i=1}^{n}b_{i}X_{i}+c\,$

has, with respect to a suitable

orthonormal basis (with respect to the standard inner product, and allowing translations), the form (${}k\geq n$ and ${}r_{i}\neq 0$)

${}F=\sum _{1\leq i\leq k}r_{i}U_{i}^{2}+s\,,$

or the form (${}k\leq n-1$ and ${}r_{i}\neq 0$)

${}F=\sum _{1\leq i\leq k}r_{i}U_{i}^{2}+sU_{k+1}\,.$

Proof

We consider the square matrix

${}M={\left(\alpha _{ij}\right)}_{1\leq i,j\leq n}\,,$

where

${}\alpha _{ij}={\begin{cases}a_{ij}{\text{ for }}i=j\,,\\{\frac {a_{ij}}{2}}{\text{ for }}i<j\,,\\{\frac {a_{ji}}{2}}{\text{ for }}i>j\,.\end{cases}}\,$

With this, the pure-quadratic term of the polynomial has the form

$\left(X_{1},\,\ldots ,\,X_{n}\right)M{\begin{pmatrix}X_{1}\\\vdots \\X_{n}\end{pmatrix}}.$

This equation holds upon inserting any element from ${}K$ for ${}X_{i}$, and also as an equation in ${}K[X_{1},\ldots ,X_{n}]$. By definition, this matrix ${}M$ is symmetric. Because of fact, there exists a orthonormal basis ${}v_{1},\ldots ,v_{n}$ of ${}\mathbb {R} ^{n}$ such that the new Gram matrix

${B^{\text{tr}}}MB$

(describing the form with respect to the new basis, ${}B$ denotes the base change matrix) has diagonal form. Let ${}V_{1},\ldots ,V_{n}$ be the new variables with respect to the new orthonormal system; that is, the ${}V_{i}$ describe, considered as functions, the linear forms to this new basis, that is, the dual basis. In the new variables, there are no mixed quadratic terms any more; the polynomial has now the form

${}F=\sum _{1\leq i\leq k}e_{i}V_{i}^{2}+\sum _{j=1}^{n}f_{j}V_{j}+g\,,$

with a certain ${}k$ between ${}1$ and ${}n$, and ${}e_{i}\neq 0$. The summands

$e_{i}V_{i}^{2}+f_{i}V_{i}$

can be brought, by completing the square and using new variables ${}U_{i}=V_{i}+h_{i}$, to the form

$e_{i}U_{i}^{2}+g_{i}.$

Besides the pure-quadratic term, either a constant or a linear polynomial remains. In the second case, we denote this linear form by ${}U_{k+1}$.

$\Box$

The representations occurring in this theorem are called the standard form of the quadratic form. In a standard form, we only have purely-quadratic terms and at most one variable in first degree. The theorem tells us that every quadratic form can be brought, using suitable orthonormal (Cartesian) coordinates, into such a standard form. Regarding the vanishing set, such a coordinate transformation means that we apply an affine-linear isometry.

The coefficients ${}r_{i}$ in the two standard forms occurring in fact are just the eigenvalues ${}\neq 0$ from the matrix ${}M$ defined in the proof. Therefore, apart from the linear and the constant term, we can write down the standard form directly, when we know the eigenvalues.

Remark

A quadratic form in standard form

$\sum _{1\leq i\leq k}r_{i}U_{i}^{2}+s\,\,{\text{ or }}\,\,\sum _{1\leq i\leq k}r_{i}U_{i}^{2}+sU_{k+1}$

in the sense of fact can be simplified further, if we allow distortions. In the new coordinates

${}Z_{i}={\sqrt {\vert {r_{i}}\vert }}U_{i}\,$

or

${}U_{i}={\frac {1}{\sqrt {\vert {r_{i}}\vert }}}Z_{i}\,$

for ${}i=1,\ldots ,k$, the quadratic form has a representation of the form

$\sum _{1\leq i\leq k}\pm Z_{i}^{2}+s\,\,{\text{ or }}\,\,\sum _{1\leq i\leq k}\pm Z_{i}^{2}+sZ_{k+1},$

the coefficients are ${}1$ or ${}-1$. This is called the normalized standard form of the quadratic form. By swapping of the variables, we may achieve that the first variables have the coefficient ${}1$, and the later variables have coefficient ${}-1$. In doing these transformations, the vanishing sets are distorted. For example, an ellipse might become a circle, or a parabola might be compressed. Since the vanishing set does not change by multiplying the form with ${}-1$, we may also assume that the number variables with coefficient ${}1$ is at least the number of variables with coefficient ${}-1$.

Example

We consider the quadratic polynomial

${}F=3X^{2}-4XY+5Y^{2}+6X+2Y-7\,.$

We have to diagonalize the matrix

${}M={\begin{pmatrix}3&-2\\-2&5\end{pmatrix}}\,.$

The characteristic polynomial is

${}(X-3)(X-5)-4=X^{2}-8X+11=(X-4)^{2}-5\,.$

Therefore, the eigenvalues are

$x_{1}={\sqrt {5}}+4{\text{ and }}x_{2}=-{\sqrt {5}}+4.$

Eigenvectors are

${\begin{pmatrix}-2\\{\sqrt {5}}+1\end{pmatrix}}\,\,{\text{ and }}\,\,{\begin{pmatrix}2\\{\sqrt {5}}-1\end{pmatrix}}.$

Hence,

${\frac {1}{\sqrt {10+2{\sqrt {5}}}}}{\begin{pmatrix}-2\\{\sqrt {5}}+1\end{pmatrix}}\,\,{\text{ and }}\,\,{\frac {1}{\sqrt {10-2{\sqrt {5}}}}}{\begin{pmatrix}2\\{\sqrt {5}}-1\end{pmatrix}}.$

is an orthonormal basis consisting of eigenvectors.

Example

We make a list of real quadratic polynomials in the three variables ${}X,Y$ and ${}Z$, together with their corresponding vanishing sets, where we restrict the coefficients to ${}0,1,-1$. Moreover, we consider only such polynomials where all variables occur and the vanishing set is not empty.

• ${}Y^{2}+X^{2}-Z\,$ ${}\,$ ${}\,$ the vanishing set is a paraboloid.
• ${}Y^{2}-X^{2}-Z\,$ ${}\,$ ${}\,$ the vanishing set is a saddle surface.
• ${}X^{2}+Y^{2}+Z^{2}$ ${}\,$ ${}\,$ the only solution is the point ${}(0,0,0)$, the vanishing set is just one point.
• ${}X^{2}+Y^{2}+Z^{2}-1$ ${}\,$ ${}\,$ the vanishing set is a sphere, that is, the surface of a ball.
• ${}X^{2}+Y^{2}-Z^{2}$ ${}\,$ ${}\,$ the vanishing set is the solution set of the equation
${}Z^{2}=X^{2}+Y^{2}$. This is a (double)-cone.
• ${}X^{2}+Y^{2}-Z^{2}-1$ ${}\,$ ${}\,$ the vanishing set is a one-sheeted hyperboloid.
• ${}X^{2}+Y^{2}-Z^{2}+1$ ${}\,$ ${}\,$ the vanishing set is a two-sheeted hyperboloid.

Example

We consider the quadratic form

${\frac {3}{2}}x^{2}+2y^{2}+2xy-2yz.$

The corresponding symmetric matrix is

${\begin{pmatrix}{\frac {3}{2}}&1&0\\1&2&-1\\0&-1&0\end{pmatrix}}.$

We want to find an orthonormal basis of ${}\mathbb {R} ^{3}$ such that, with respect to the new basis, the form is described by a diagonal matrix. For this, we have to determine the eigenvalues (principal values) of the matrix. The characteristic polynomial of the matrix is

${}{\begin{aligned}\det {\begin{pmatrix}X-{\frac {3}{2}}&-1&0\\-1&X-2&1\\0&1&X\end{pmatrix}}&={\left(X-{\frac {3}{2}}\right)}{\left(X^{2}-2X-1\right)}-X\\&=X^{3}-{\frac {7}{2}}X^{2}+X+{\frac {3}{2}}\\&=(X-1){\left(X^{2}-{\frac {5}{2}}X-{\frac {3}{2}}\right)}\\&=(X-1)(X-3){\left(X+{\frac {1}{2}}\right)};\end{aligned}}$

hence, the eigenvalues are

$1,3,-{\frac {1}{2}}.$

The corresponding principal axes can be determined in the following way.

For ${}x=1$, the kernel of the matrix

${\begin{pmatrix}-{\frac {1}{2}}&-1&0\\-1&-1&1\\0&1&1\end{pmatrix}}$

equals ${}{\begin{pmatrix}2\\-1\\1\end{pmatrix}}$, a normalized generator is

${\begin{pmatrix}{\frac {2}{\sqrt {6}}}\\-{\frac {1}{\sqrt {6}}}\\{\frac {1}{\sqrt {6}}}\end{pmatrix}}.$

For ${}x=3$, the kernel of the matrix

${\begin{pmatrix}{\frac {3}{2}}&-1&0\\-1&1&1\\0&1&3\end{pmatrix}}$

equals ${}{\begin{pmatrix}2\\3\\-1\end{pmatrix}}$, a normalized generator is

${\begin{pmatrix}{\frac {2}{\sqrt {14}}}\\{\frac {3}{\sqrt {14}}}\\-{\frac {1}{\sqrt {14}}}\end{pmatrix}}.$

For ${}x=-{\frac {1}{2}}$, the kernel of the matrix

${\begin{pmatrix}-2&-1&0\\-1&-{\frac {5}{2}}&1\\0&1&-{\frac {1}{2}}\end{pmatrix}}$

equalsh ${}{\begin{pmatrix}-{\frac {1}{2}}\\1\\2\end{pmatrix}}$, a normalized generator is

${\begin{pmatrix}-{\frac {1}{4{\sqrt {21}}}}\\{\frac {1}{2{\sqrt {21}}}}\\{\frac {1}{\sqrt {21}}}\end{pmatrix}}.$

We denote these eigenvectors by ${}u_{1},u_{2},u_{3}$, they form an orthonormal basis. In the new coordinates ${}y_{1},y_{2},y_{3}$, given by the new orthonormal basis, the quadratic form is written as

$y_{1}^{2}+3y_{2}^{2}-{\frac {1}{2}}y_{3}^{2}.$

This is clear already just by looking at the eigenvalues; for this, the computation of the eigenvectors not necessary.

Between the two bases, we have the relation

${}{\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}={\begin{pmatrix}{\frac {2}{\sqrt {6}}}&-{\frac {1}{\sqrt {6}}}&{\frac {1}{\sqrt {6}}}\\{\frac {2}{\sqrt {14}}}&{\frac {3}{\sqrt {14}}}&-{\frac {1}{\sqrt {14}}}\\-{\frac {1}{4{\sqrt {21}}}}&{\frac {1}{2{\sqrt {21}}}}&{\frac {1}{\sqrt {21}}}\end{pmatrix}}{\begin{pmatrix}e_{1}\\e_{2}\\e_{3}\end{pmatrix}}\,.$

Due to fact, we have the relation

${}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}{\frac {2}{\sqrt {6}}}&{\frac {2}{\sqrt {14}}}&-{\frac {1}{4{\sqrt {21}}}}\\-{\frac {1}{\sqrt {6}}}&{\frac {3}{\sqrt {14}}}&{\frac {1}{2{\sqrt {21}}}}\\{\frac {1}{\sqrt {6}}}&-{\frac {1}{\sqrt {14}}}&{\frac {1}{\sqrt {21}}}\end{pmatrix}}{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}={\begin{pmatrix}{\frac {2}{\sqrt {6}}}y_{1}+{\frac {2}{\sqrt {14}}}y_{2}-{\frac {1}{4{\sqrt {21}}}}y_{3}\\-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\\{\frac {1}{\sqrt {6}}}y_{1}-{\frac {1}{\sqrt {14}}}y_{2}+{\frac {1}{\sqrt {21}}}y_{3}\end{pmatrix}}\,$

between the coordinates ${}x_{1},x_{2},x_{3}$ (the dual bases of the standard basis; these coordinates were denote $x,y,z$ in the beginning) and the coordinates ${}y_{1},y_{2},y_{3}$ of the new orthogonal basis.