Quadrics

Orthogonal and symmetric maps: Applications of symmetric maps

Quadrics

A quadratic form on a vector space can be regarded as a homogeneous, second-degree polynomial in the coordinates of a vector relative to a fixed chosen basis of the vector space. An ordinary quadratic polynomial is the sum of a quadratic form, a linear map, and a constant function.

A quadratic polynomial function $p$ on a vector space $V$ is the sum of a quadratic form $q$ unequal to $0$ , a linear function $L:V\to\mathbb{R}$ and a constant function $c$ .

Let $\alpha$ be a basis for $V$ . Then the function $p$ as above is determined by
$\begin{array}{rcl}p(\alpha^{-1}(\vec{x})) &=& (x_1\cdots x_n)\ A\left(\,\begin{array}{c} x_1\\ \vdots\\ x_n \end{array}\,\right)\ +(l_1 \cdots l_n)\ \left(\,\begin{array}{c} x_1\\ \vdots\\ x_n \end{array}\,\right)\ +c \\&=& \dotprod{\vec{x}}{(A\,\vec{x})} +\dotprod{\vec{l}}{\vec{x}} +c\end{array}$ for $\vec{x}$ in $\mathbb{R}^n$ , where $A$ is the matrix of $q$ with respect to $\alpha$ and $\vec{l} = \rv{l_1,\ldots,l_n}$ satisfies $L(\vec{x}) =\dotprod{\vec{l}}{\vec{x}}$ .

The quadric determined by $p$ is the set of vectors $\vec{v}$ of $V$ that satisfy $p(\vec{v}) = 0$ .

Two quadrics are called congruent if one can be obtained from the other by application of an isometry. If $K$ is an isometry, then the quadric determined by $p$ as above is congruent to the quadric determined by $p\circ K$ .

Often we write the coordinates of the vector as arguments of $p$ and $q$ rather than the vector itself. Thus, we do not make a distinction between $p(x,y)$ and $p(\rv{x,y})$ if $\rv{x,y}$ is a vector of $\mathbb{R}^2$ .

3-dimensional example The quadratic polynomial $p$ on $\mathbb{R}^3$ determined by $p(x,y,z) = 2x^2 - 4xy+4xz-3y^2-3z^2+5x-7y+z-1$ can also be written as $p(x,y,z)= \matrix{x & y & z}\, \matrix{2&-2&2\\-2&-3&0\\ 2&0&-3}\, \matrix{x\\ y\\ z} +\matrix{5&-7&1}\,\matrix{x\\ y\\ z} -1$

A quadric is also called quadratic (hyper) surface. If the vector space has dimension $2$ , this is a quadratic curve of the plane. If the dimension is equal to $3$ , we also speak of a quadratic surface.

If $R$ is a quadric, then, up to a constant factor, there is a unique quadratic polynomial function determining $R$ . This can be proved by showing that, up to a constant factor, the polynomial function is fully determined by the fact that it has value $0$ on sufficiently many points of $\mathbb{R}^n$ . A concrete lower limit on this number is $\frac12\cdot (n+1)\cdot (n+2)$ equal to the dimension of the linear space of all quadratic forms on $\mathbb{R}^{n+1}$ . In particular, a quadratic curve of the plane is determined by $\frac12\cdot 3\cdot 4=6$ of its points with the property that no three of them lie on a single line.

Let $Q$ be the quadric determined by $p$ and $K$ an invertible linear map $V\to V$ . Then $\vec{v}$ belongs to $Q$ if and only if $K^{-1}(\vec{v})$ is on the quadric of $p\circ K$ . After all, $(p\circ K)(K^{-1}\vec{v}) = p(K(K^{-1}\,\vec{v}))= p(\vec{v})$ so $(p\circ K)(\vec{v})$ is equal to zero precisely if $p(K\vec{v}) = 0$ . This shows that the image under $K$ of the quadric determined by the standard polynomial $p\circ K$ is the quadric of $p$ .

The converse of the statement regarding congruence is also true. If $R$ is a quadric that is congruent to $Q$ , then, by definition, there is an isometry $L$ which transforms $R$ into $Q$ , so $R$ is the quadric determined by $p\circ L$ .

2-dimensional example Consider the quadratic form $q(x,y) =2x^2-4xy+5y^2=q\left(\beta(\rv{x,y})\right) = \frac15(2x+y)^2+\frac65(x-2y)^2$ on $\mathbb{R}^2$ where $\beta= \basis{\frac{1}{\sqrt{5}}\rv{2,1},\frac{1}{\sqrt{5}}\rv{1,-2}}$ The expression as a linear combination of two squares is not unique. For example, by completing the square, the function $q$ can be written as a sum of quadratic functions as follows: $q(x,y) = 2(x-y)^2 +3y^2$ In terms of matrices this equation can be written as $q(x,y) = q'(N(\rv{x,y}))$ where $q'(x,y) = 2x^2+3y^2\quad\text{ and }\quad N = \matrix{1&-1\\ 0&1}$ By replacing $\rv{x,y}$ left and right by $N^{-1}(\rv{x,y})$ , we find $q\circ\alpha^{-1} (\rv{x,y}) = q'(x,y)=2x^2+3y^2$ where $\alpha^{-1}$ consists of the columns of $N^{-1} = \matrix{1&1\\ 0&1}$ Therefore, the matrix of the quadratic form with respect to the basis $\alpha$ is also diagonal. But the matrix $N$ is not orthogonal. Therefore, we cannot conclude that the quadric determined by $q(x,y)$ is congruent to the ellipse determined by $q'(x,y)$ . There is no isometry that transforms the ellipse determined by $q'$ into the ellipse determined by $q$ since the lengths of the axes (that is, line segments connecting pairs of points at largest distance on the ellipse) differ between $q$ and $q'$ .

Theorem Quadratic forms and symmetric matrices says every quadratic form can be put in diagonal form by means of an orthogonal map. Here we show that every quadratic polynomial function can be brought in standard form by use of an isometry.

Let $p$ be a quadratic polynomial function on $\mathbb{R}^n$ . Then there are an isometry $K:\mathbb{R}^n\to\mathbb{R}^n$ , an index $r\le n$ , and real numbers $l_1,\ldots,l_{r+1}$ such that $l_1$ is unequal to $0$ , and

$p(K(\vec{x})) =\begin{cases}\displaystyle\sum_{i=1}^r l_i\cdot x_i^2 +l_{r+1}\cdot x_{r+1} &\\ \phantom{xxx}\text{ or}&\\\displaystyle\sum_{i=1}^r l_i\cdot x_i^2 +l_{r+1} &\end{cases}$

where $\vec{x} = \rv{x_1,\ldots,x_n}$ .

The quadratic polynomial function $p$ can be written as follows as a function of a column vector $\vec{x}$
$p(\vec{x}) = \vec{x}^\top\, A\, \vec{x}+\vec{l}^\top\, \vec{x}+c$ Here, $A$ is a symmetric matrix, $\vec{l}$ is a column vector, and $c$ is a real number. We view the expression $\vec{l}^\top\, \vec{x}$ as the matrix product of a $(1\times n)$ -matrix with an $(n\times 1)$ -matrix, which results in a real number; it is a different way of writing the inner product $\dotprod{\vec{l}}{\vec{x}}$ . Similarly, $\vec{x}^\top\, A\, \vec{x}$ is equal to the inner product $\dotprod{\vec{x}}{(A\,\vec{x})}$ .

Let $\alpha =\basis{\vec{a}_1,\ldots ,\vec{a}_n}$ be an orthonormal basis of eigenvectors of $A$ with eigenvalues $\lambda_1, \ldots,\lambda_n$ . Here we choose the order in such a way that the eigenvalues are equal to $0$ at the end. As a consequence, there is an index $r\le n$ with $\lambda_i\ne0$ for $i\le r$ and $\lambda_i=0$ for $i\gt r$ . We indicate the coordinates of $\vec{x} = \rv{x_1, \ldots ,x_n}$ with respect to $\alpha$ by $\vec{x}' = \rv{x_1',\ldots, x_n'}$ . The correspondence is
$\left(\,\begin{array}{c} x_1\\ \vdots\\ x_n \end{array}\,\right)=B\ \left(\,\begin{array}{c} x_1'\\ \vdots\\ x_n' \end{array}\,\right)$ where the columns of $B$ are the vectors $\vec{a}_1,\ldots ,\vec{a}_n$ , so $B = {}_\varepsilon I_\alpha$ . In terms of vectors: $\vec{x} =B\, \vec{x}'$ Substitution in the function rule for $p(\vec{x})$ gives $\begin{array}{rcl}p(\vec{x}) &=& (\vec{x}' )^\top B^\top\,A\,B\,\vec{x}' +\vec{l}^\top\,B\,\vec{x}'+c\\&=&( \vec{x}' )^\top\, D\,\vec{x}' +(\vec{l}\,')^\top\,\vec{x}'+c\\ &=& \lambda_1x_1'^2+\cdots + \lambda_nx_n'^2+l_1'x_1'+\cdots +l_n'x_n'+c\end{array}$ Here, $D=B^\top\,A\,B$ is the $(n\times n)$ -diagonal matrix with the eigenvalues $\lambda_1,\ldots,\lambda_n$ on the diagonal and $\vec{l}\,'$ is given by
$\vec{l}\,' =\left(\vec{l}^\top\, B \right)^\top=B^\top\,\vec{l}$ Thus, the components of $\vec{l}\,'$ are the $\alpha$ -coordinates of the vector $\vec{l}$ .

If $i\leq r$ , then we can eliminate the linear term with $x_i'$ by completing the square. Here we proceed by using translations: Let $\vec{a}$ be a column vector in $\mathbb{R}^n$ , and write $\vec{x}\,'' = T_{-\vec{a}}(\vec{x}')$ so $\vec{x}\,' = \vec{x}\,''+\vec{a}$ . Substituting this expression for $\vec{x}\,'$ in $p(\vec{x})$ gives $\begin{array}{rcl}p(\vec{x}) &=& \left(\vec{x}\,''+\vec{a}\right)^\top \,D\,(\vec{x}\,''+\vec{a}) +(\vec{l}\,')^\top\,(\vec{x}\,''+\vec{a})+c\\ &&\color{blue}{\vec{x}\,' = \vec{x}\,''+\vec{a}\text{ substituted in }p(\vec{x})=(\vec{x}' )^\top\, D\,\vec{x}' +(\vec{l}\,')^\top\,\vec{x}'+c}\\&=& (\vec{x}\,'')^\top \,D\,\vec{x}\,''+(\vec{x}\,'')^\top\,D\,\vec{a}+\vec{a}^\top\,D\,\vec{x}\,''+\vec{a}^\top\, D \, \vec{a} +(\vec{l}\,')^\top\,\vec{x}\,''+(\vec{l}\,')^\top\,\vec{a}+c\\&&\color{blue}{\text{brackets expanded}}\\&=& (\vec{x}\,'')^\top \,D\,\vec{x}\,''+\left(\vec{a}^\top\,D^\top\,\vec{x}\,''\right)^\top+\vec{a}^\top\,D\,\vec{x}\,''+\vec{a}^\top\, D \, \vec{a} +(\vec{l}\,')^\top\,\vec{x}\,''+(\vec{l}\,')^\top\,\vec{a}+c\\&&\color{blue}{\text{computational rule }\left(\vec{a}^\top\,D^\top\,\vec{x}\,''\right)^\top=(\vec{x}\,'')^\top\,D\,\vec{a}}\\&=& (\vec{x}\,'')^\top \,D\,\vec{x}\,'' +2\vec{a}^\top\,D\,\vec{x}\,''+\vec{a}^\top\, D \, \vec{a} +(\vec{l}\,')^\top\,\vec{x}\,''+(\vec{l}\,')^\top\,\vec{a}+c\\&&\color{blue}{D^\top=D\text{ and }\left(\vec{a}^\top\,D\,\vec{x}\,''\right)^\top=\vec{a}^\top\,D\,\vec{x}\,''\text{ since this is a scalar}}\\ &=& (\vec{x}\,'')^\top \,D\,\vec{x}\,''+\left(2D\,\vec{a}+\vec{l}\,'\right)^\top\,\vec{x}\,''+c'\\&&\color{blue}{\text{rewritten with }c' = \vec{a}^\top\, D \, \vec{a} +(\vec{l}\,')^\top\,\vec{a}+c}\end{array}$ In order to eliminate the terms linear in $x_i$ for $i\le r$ , we choose $\vec{a} = -\frac12\cdot D'\vec{l}\,'$ where $D'$ is the diagonal matrix with $\lambda_i^{-1}$ as its $i$ -th diagonal entry for $i\le r$ and zeros elsewhere. So, on the subspace perpendicular to the kernel of $D$ , the matrix $D'$ is the inverse of $D$ , and $\vec{a}^\top=-\frac12 \cdot \rv{\lambda_1^{-1}l_1',\ldots,\lambda_r^{-1}l_r',0,\ldots,0}$ The result is $\begin{array}{rcl}p(\vec{x}) &=& (\vec{x}\,'')^\top \,D\,\vec{x}\,'' +\left(\vec{l}\,''\right)^\top \,\vec{x}\,''+c'\end{array}$ where $\vec{l}\,''$ is the vector with zeros in the first $r$ coordinates and $l_i''=l_i'$ for $i=r+1,\ldots,n$ .

If $\vec{l}\,''=\vec{0}$ , then the function rule for $p$ is as required, that is, as in the second case with $l_{r+1} = c'$ . To see this, we choose $L = B\,T_{\vec{a}}$ so $\vec{x}=L\, (\vec{x}\,'')$ and $p(L(\vec{x}\,'')) = p(\vec{x}) = (\vec{x}\,'')^\top \,D\,\vec{x}\,'' +l_{r+1}$ If we replace the argument $\vec{x}\,''$ on both sides by $\vec{x}$ , we find the required formula.

Assume, therefore, that $\vec{l}\,''$ is not equal to the zero vector. In the rest of the proof we will only consider isometries of $\mathbb{R}^n$ which fix the first $r$ coordinates. So they will leave the following linear subspace invariant: $W = \linspan{\vec{e}_1,\ldots,\vec{e}_r}^\perp=\linspan{ \vec{e}_{r+1},\ldots,\vec{e}_n}$ Because the first $r$ coordinates of $\vec{l}\,''$ are equal to zero, this vector lies in $W$ .

Put $l_{r+1} = \norm{\vec{l}\,''}$ . There is an orthogonal map $S:\mathbb{R}^n\to\mathbb{R}^n$ which fixes each vector in $W^\perp=\linspan{\vec{e}_1,\ldots,\vec{e}_r}$ and maps $l_{r+1}\vec{e}_{r+1}$ onto $\vec{l}\,''$ . An example is the orthogonal reflection $S = S_{l_{r+1}\vec{e}_{r+1}-\vec{l}\,''}$ . Because $S$ is orthogonal, according to property 5 of orthogonal maps it also leaves $W$ invariant, such that we find for an arbitrary vector $\vec{y}$ in $\mathbb{R}^n$ : $\begin{array}{rcll}S\,D\,\vec{y}&=&S\,D\left(\vec{w}+\vec{m}\right)&\color{blue}{\text{direct sum decomposition with }\vec{w}\in W\text{ and }\vec{m}\in W^\perp}\\&=&S\,D\,\vec{m}&\color{blue}{\vec{w}\in W=\ker{D}}\\&=&D\,\vec{m}&\color{blue}{D\,\vec{m}\in W^\perp\text { and }S\text{ fixes }W^\perp}\\&=&D\left(S\,\vec{w}+\vec{m}\right)&\color{blue}{S\,\vec{w}\in W=\ker{D}}\\&=&D\,S\left(\vec{w}+\vec{m}\right)&\color{blue}{\vec{m}\in W^\perp\text{ and }S\text{ fixes }W^\perp}\\&=&D\,S\,\vec{y}&\color{blue}{\vec{y}=\vec{w}+\vec{m}}\\\end{array}$ Below, we use this result in the form $S^{-1}\,D=D\,S^{-1}$ . Write $\vec{x}\,''' = S^{-1}\vec{x}\,''$ Then we have $\vec{x}\,''' = \vec{x}\,''$ for all $\vec{x}''$ in $W^\perp$ , such that $\begin{array}{rcl}p(\vec{x}) &=& (\vec{x}\,'')^\top \,D\,\vec{x}\,'' +\left(\vec{l}\,''\right)^\top \,\vec{x}\,''+c'\\&&\color{blue}{\text{previously found formula}}\\&=&(\vec{x}\,'')^\top\left(S^{-1}\right)^\top\,S^{-1}\,D\,\vec{x}\,'' +\left(\vec{l}\,''\right)^\top\left(S^{-1}\right)^\top\,S^{-1} \,\vec{x}\,''+c'\\&&\color{blue}{\left(S^{-1}\right)^\top\,S^{-1}=I_n\text{ because }S\text{ is orthogonal}}\\&=& (S^{-1}\vec{x}\,'')^\top\,\,D\,\left(S^{-1}\vec{x}\,''\right)+\left(S^{-1}\vec{l}\,''\right)^\top \,(S^{-1}\vec{x}\,'')+c'\\&&\color{blue}{\text{computational rule }\left(X\,Y\right)^\top=Y^\top\,X^\top\text{ and }S^{-1}\,D=D\,S^{-1}}\\&=& (\vec{x}\,''')^\top \,D\,\vec{x}\,''' +\left(l_{r+1}\vec{e}_{r+1}\right)^\top \,\vec{x}\,'''+c'\\&&\color{blue}{\vec{x}\,''' = S^{-1}\vec{x}\,''\text{ and }S\left(l_{r+1}\vec{e}_{r+1}\right)=\vec{l}\,''}\end{array}$ Finally, we choose the vector $\vec{b} = c'\cdot l_{r+1}^{-1}\,\vec{e}_{r+1}$ and we write $\vec{x}\,'''' = T_{-\vec{b}}\,\vec{x}\,'''$ so $\vec{x}\,''' = \vec{x}\,'''' - \vec{b}$ and $\begin{array}{rcl}p(\vec{x}) &=&(\vec{x}\,''')^\top \,D\,\vec{x}\,''' +\left(l_{r+1}\vec{e}_{r+1}\right)^\top \,\vec{x}\,'''+c'\\&&\color{blue}{\text{formula from above}}\\&=&(\vec{x}\,''''- \vec{b})^\top \,D\,\left(\vec{x}\,''''- \vec{b}\right)+\left(l_{r+1}\vec{e}_{r+1}\right)^\top \,(\vec{x}\,''''-\vec{b})+c'\\&&\color{blue}{\vec{x}\,''' = \vec{x}\,'''' - \vec{b}}\\&=& (\vec{x}\,'''')^\top \,D\,\vec{x}\,'''' +\left(l_{r+1}\vec{e}_{r+1}\right)^\top \,\vec{x}\,''''-\left(l_{r+1}\vec{e}_{r+1}\right)^\top \,\vec{b}+c'\\&&\color{blue}{\vec{b}^\top\,D=\vec{0}^\top\text{ and }D\,\vec{b}=\vec{0}\text{ because }\vec{b}\in W=\ker{D}}\\&=& (\vec{x}\,'''')^\top \,D\,\vec{x}\,'''' +\left(l_{r+1}\vec{e}_{r+1}\right)^\top \,\vec{x}\,''''-c'\left(l_{r+1}\vec{e}_{r+1}\right)^\top \, l_{r+1}^{-1}\vec{e}_{r+1}+c'\\&&\color{blue}{\vec{b} = c'\cdot l_{r+1}^{-1}\,\vec{e}_{r+1}\text{ substituted}}\\&=& (\vec{x}\,'''')^\top \,D\,\vec{x}\,'''' +l_{r+1}\cdot \vec{e}_{r+1}^\top \,\vec{x}\,''''\\&&\color{blue}{\text{last two terms cancel thanks to our choice for }\vec{b}}\\&=& (\vec{x}\,'''')^\top \,D\,\vec{x}\,''''+l_{r+1}\cdot {x''''}_{r+1}\end{array}$ This shows that we arrive at the function rule of the first case.

We verify that, also in the first case, the result can be formulated as indicated in the theorem. Composing all transformations involved, we see that $K= B\,T_{\vec{a}}\,S\,T_{\vec{b}}$ is an isometry that satisfies $\vec{x} = K(\vec{x}\,'''')$ : $\vec{x} = B\, \vec{x}\,' = B\,T_{\vec{a}}\vec{x}\,''=B\,T_{\vec{a}}\,S\,\vec{x}\,''' = B\,T_{\vec{a}}\,S\,T_{\vec{b}}\,\vec{x}\,''''=K(\vec{x}\,'''')$ Using this in the function rule for $p$ , we find $p(K(\vec{x}\,'''')) = (\vec{x}\,'''')^\top \,D\,\vec{x}\,'''' +l_{r+1}\cdot {x''''}_{r+1}$ Finally, we replace $\vec{x}\,''''$ by $\vec{x}$ , and arrive at $\begin{array}{rcl}p(K(\vec{x})) &=& \vec{x}^\top \,D\,\vec{x} +l_{r+1}\cdot {x}_{r+1}\\ & =&\displaystyle\sum_{i=1}^r l_i\cdot x_i^2 +l_{r+1}\cdot x_{r+1}\end{array}$

The proof also provides a method for bringing a quadric defined by $p(\vec{x}) = \vec{x}^\top\, A\, \vec{x}+\vec{l}^\top\, \vec{x}+c$ into standard form by means of isometries:

Step 1: Find an orthonormal basis $\alpha$ of $\mathbb{R}^n$ consisting of eigenvectors of $A$ . This gives the orthogonal matrix $B = {}_\varepsilon I_\alpha$ such that $B^\top \, A\ B$ is in diagonal form.
Step 2: Cancel with the aid of a translation $T_{\vec{a}}$ the linear terms of variables that also occur in the homogeneous quadratic part.

If there remains no linear term, then $K = B\,T_{\vec{a}}$ is an isometry as required. Otherwise, we carry out two more steps:

Step 3: Rewrite the linear part with the aid of an orthogonal transformation to a constant multiple of $x_{r+1}$ , where $r$ is the rank of $A$ . The transformation must fix the first $r$ basis vectors. The orthogonal reflection $S = S_{\norm{\vec{l}\,''}\cdot \vec{e}_{r+1}-\vec{l}\,''}$ satisfies, where $\vec{l}\,''$ is the vector of the coefficients of the linear terms.
Step 4: Cancel the constant term by means of a translation $T_{\vec{b}}$ .

Now $K = B\,T_{\vec{a}}\,S\,T_{\vec{b}}= B\, S\,T_{S\vec{a}+\vec{b}}$ is an isometry with the property that $(p\circ K)(\vec{x})$ is in standard form. Examples of the working of this algorithm are given at the bottom of the page.

We consider the consequences of the theorem for quadratic curves. Up to congruence and multiplication by a nonzero constant, there are nine cases for the equation of the curve in terms of $\rv{x,y}$ , where the polynomial function is scaled such that the coefficient of $x^2$ equals $1$ . We denote the coefficient of $y^2$ by $b$ , the coefficient of $y$ by $c$ , and the constant term by $d$ :

$\begin{array}{lrccl}\text{equation}&&\text{geometric figure}&&\text{restrictions on parameters}\\ \hline x^2+y^2 = d&\phantom{xxx}&\text{circle}&\phantom{xxx}&d\gt0\\ x^2+by^2 = d&\phantom{xxx}&\text{ellipse but not a circle}&\phantom{xxx}&b,d\gt0, b\ne1\\ x^2+by^2 = d&\phantom{xxx}&\text{hyperbola}&\phantom{xxx}&b\lt0,d\ne0\\ x^2+cy = 0&\phantom{xxx}&\text{parabola}&\phantom{xxx}&c\ne0\\ x^2+by^2=0&\phantom{xxx}&\text{two intersecting lines}&\phantom{xxx}&b\lt0\\ x^2 = d&\phantom{xxx}&\text{two parallel lines}&\phantom{xxx}&d\gt0\\ x^2 = 0&\phantom{xxx}&\text{a single line}&\phantom{xxx}& \\ x^2+by^2 = 0&\phantom{xxx}&\text{a single point}&\phantom{xxx}& b\gt0\\ x^2+by^2 = d&\phantom{xxx}&\text{empty set}&\phantom{xxx}&b\ge0,\,d\lt0\end{array}$

In general, the fact that the degree equals 2 means that every line intersects the curve in two points. Exceptions are the single line, the point, and the empty set. In the case of one line, this rule is often made valid by speaking of two overlapping lines. In the other two cases, the two points can be found after extension of the plane to a complex inner product space.

The type of geometric figure, where we count a circle as being an ellipse, is also invariant under affine transformations (that is, compositions of translations and invertible linear maps). The precise coefficients in the standard form will not be preserved, but the signs of the quadratic terms will be.

The shape and size of a quadric remains unchanged under an isometry $L$ . The isometry $L$ only changes its position in space. The quadric determined by $p$ does not change when $p(\vec{x})$ is multiplied by a nonzero scalar.

The second step of the proof as well as the algorithm is in fact a completion of the square. The translation along $\vec{a}$ can be described as follows: If $i\leq r$ , we write
$\lambda_ix_i'^2+b_i'x_i'=\lambda_i\Big( x_i'^2+\frac{b_i'}{\lambda_i}x_i\Big)\ =\lambda_i \Big( x_i'+\frac{b_i'}{2\lambda_i}\Big)^2-\frac{b_i'^2}{4\lambda_i^2}$ For all $i$ with $i\leq r$ we replace $x_i'$ by $x_i''-\frac{b_i'}{2\lambda_i}$ .

The origin is a special point for the quadratic polynomial functions in standard form. It is a sort of center of gravity and the point of intersection of the main axes (read: the $1$ -dimensional spans of linearly independent eigenvectors of $A$ ).

Uniqueness The standard form is not unique. For example, the ordering of diagonal entries of $A$ is not fixed (although we often order the diagonal entries according to decreasing absolute value), and the coefficient of the linear term with $x_{r+1}$ can be multiplied by $-1$ .

Consider the quadratic function on $\mathbb{R}^2$ given by
$p(x,y) = 4 x^2+4 y^2-8 y+2$ Determine a standard form $p'$ for $p$ .

$p' (x,y) =$ $4 x^2+4 y^2-2$

We write $\begin{array}{rcl}p(x,y) &=& \matrix{x&y}\, A\, \matrix{x\\ y} + L \cv{x\\ y} + 2 \\ &&\text{where }A\text{ is the matrix of the quadratic form of }p\text{:}\\ A &=& \matrix{4 & 0 \\ 0 & 4 \\ }\\&&\text{and }L\text{ the }(1\times 2)\text{-matrix of coefficients of the linear form of }p\text{: }\\ L &=& \matrix{0 & -8 \\ }\end{array}$ The characteristic polynomial of $A$ in the variable $t$ is
$p_A(t) = t^2-8 t+16= \left(t-4\right)^2$ Therefore, the eigenvalues of $A$ are
$\begin{array}{rl} 4&\text{ with multiplicity }2 \end{array}$ Since $A$ is already in diagonal form and the eigenvalues are ordered in such a way that all entries equal to $0$ are at the end, we do not need to carry out the first step of the algorithm for bringing $p$ into standard form. In order to get rid of terms that are linear in variables that also appear in quadratic terms, we use the translation vector $\vec{a} = \left[ 0 , 1 \right]$ Substitution of $\rv{x,y}$ by $T_{\vec{a}}\rv{x,y}$ gives
$\begin{array}{rcl}(p\circ T_{\vec{a}})(x,y) &=& p(\rv{x,y} +\vec{a})\\ &=&\displaystyle p\left(\left[ x , y+1 \right] \right)\\ &=& \displaystyle 4 x^2+4 y^2-2 \end{array}$ We now come to the third step of the algorithm. There is no linear term, so the polynomial function needs no further adjustment. We have arrived at a standard form:
$\begin{array}{rcl}(p\circ T_{\vec{a}})(x,y) &=& \displaystyle 4 x^2+4 y^2-2 \end{array}$

The quadric given by $p(x,y) = 0$ can be described even simpler by dividing $p'(x,y)$ by $2$ . We conclude that the quadric of $p$ is congruent to the set of points $\rv{x,y}$ satisfying $2 x^2+2 y^2-1 =0$ The isometry that transforms the quadric of $p$ into the quadric of $p'$ is
$\begin{array}{rcl} \left(\,T_{\vec{a}}\, \, \right)^{-1} &=&\displaystyle \,T_{-\vec{a}}\, \\ &=&\displaystyle T_{\left[ 0 , -1 \right] } \end{array}$ The quadric of $2 x^2+2 y^2-1$ , and therefore also of $4 x^2+4 y^2-8 y+2$ and its standard form $4 x^2+4 y^2-2$ , is a circle.

New example