MA20216: Algebra 2A

6.3 Application: Quadratic forms

Convention. We continue working with a field $\F $ where $1+1\neq 0$.

We can construct a function on $V$ from a bilinear form $B$ (which is a function on $V\times V$).

Definition. A quadratic form on a vector space $V$ over $\F $ is a function $Q:V\to \F $ of the form
$\seteqnumber{0}{6.}{1}$
\begin{equation*} Q(v)=B(v,v), \end{equation*}

for all $v\in V$, where $B:V\times V\to \F $ is a symmetric bilinear form.

Remark. For $v\in V$ and $\lambda \in \F $, $Q(\lambda v)=B(\lambda v,\lambda v)=\lambda ^2Q(v)$ so $Q$ is emphatically not a linear function!

Examples. Here are two quadratic forms on $\F ^3$:
- (1) $Q(x)=x_1^2+x_2^2-x_3^2=B_A(x,x)$ where
  $\seteqnumber{0}{6.}{1}$
  \begin{equation*} A= \begin{pmatrix} 1&0&0\\0&1&0\\0&0&-1 \end {pmatrix}. \end{equation*}
- (2) $Q(x)=x_1x_2=B_A(x,x)$ where
  $\seteqnumber{0}{6.}{1}$
  \begin{equation*} A= \begin{pmatrix} 0&\half &0\\\half &0&0\\0&0&0 \end {pmatrix}. \end{equation*}

We can recover the symmetric bilinear form $B$ from its quadratic form $Q$:

Lemma 6.8. Let $Q:V\to \F $ be a quadratic form with $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Then
$\seteqnumber{0}{6.}{1}$
\begin{equation*} B(v,w)=\half \bigl (Q(v+w)-Q(v)-Q(w)\bigr ), \end{equation*}

for all $v,w\in V$.

$B$ is called the polarisation of $Q$.

Proof. Expand out to get
$\seteqnumber{0}{6.}{1}$
\begin{equation*} Q(v+w)-Q(v)-Q(w)=B(v,w)+B(w,v)=2B(v,w). \end{equation*}

□

Here is how to do polarisation in practice: any quadratic form $Q:\F ^n\to \F $ is of the form

\begin{equation*} Q(x)=\sum _{1\leq i\leq j\leq n}q_{ij}x_ix_j=\bx ^{T} \begin{pmatrix} q_{11}&&\half q_{ji}\\&\ddots &\\\half q_{ij}&&q_{nn} \end {pmatrix}\bx \end{equation*}

so that the polarisation is $B_A$ where

\begin{equation*} A_{ij}=A_{ji}= \begin{cases} q_{ii}&\text {if $i=j$;}\\ \half q_{ij}&\text {if $i<j$}. \end {cases} \end{equation*}

Example. Let $Q:\R ^3\to \R $ be given by
$\seteqnumber{0}{6.}{1}$
\begin{equation*} Q(x)=x_1^2+2x_2^2+2x_1x_2+x_1x_3. \end{equation*}

Let us find the polarisation $B$ of $Q$, that is, we find $A$ so that $B=B_A$: we have $q_{11}=1$, $q_{22}=2$, $q_{12}=2$ and $q_{13}=1$ with all other $q_{ij}$ vanishing so
$\seteqnumber{0}{6.}{1}$
\begin{equation*} A= \begin{pmatrix} 1&1&\half \\1&2&0\\\half &0&0 \end {pmatrix}. \end{equation*}

Definitions. Let $Q$ be a quadratic form on a finite-dimensional vector space $V$ over $\F $.

The rank of $Q$ is the rank of its polarisation.

If $\F =\R $, the signature of $Q$ is the signature of its polarisation.

What does the diagonalisation theorem mean for a quadratic form $Q$? Observe:

• Any $\alpha \in V^{*}$ can be squared to give a quadratic form: $\alpha ^2:V\to \F $ given by $\alpha ^2(v)=\alpha (v)^2$. Note that this is indeed a quadratic form with polarisation $B(v,w)=\alpha (v)\alpha (w)$.
• If $\lst {v}1n$ diagonalises the polarisation $B$ of $Q$ then $Q(\sum _i\lambda _iv_i)=\sum _iB(v_i,v_i)\lambda _i^2$ so that
$\seteqnumber{0}{6.}{1}$
\begin{equation*} Q=\sum _{i=1}^nQ(v_i)(v_i^{*})^2. \end{equation*}

That is, we have written $Q$ as a linear combination of $n$ linearly independent squares.

Let us now apply the classification results of §6.2 and summarise the situation for quadratic forms on vector spaces over our favourite fields:

Theorem 6.9. Let $Q$ be a quadratic form with rank $r$ polarisation on a finite-dimensional vector space over $\F $.
- (1) When $\F =\C $, there is a basis $\lst {v}1n$ of $V$ such that
  $\seteqnumber{0}{6.}{1}$
  \begin{equation*} Q(\sum _{i=1}^nx_iv_i)=\plst {x^2}1r. \end{equation*}
- (2) When $\F =\R $ and $Q$ has signature $(p,q)$, there is a basis $\lst {v}1n$ of $V$ such that
  $\seteqnumber{0}{6.}{1}$
  \begin{equation*} Q(\sum _{i=1}^nx_iv_i)=\plst {x^2}1p-x_{p+1}^2-\dots -x_r^2. \end{equation*}

Example. Find the signature of $Q:\R ^3\to \R $ given by
$\seteqnumber{0}{6.}{1}$
\begin{equation*} Q(x)=x_1^2+x_2^2+x_3^2+2x_1x_3+4x_2x_3. \end{equation*}

$Q$ has polarisation $B=B_A$ with
$\seteqnumber{0}{6.}{1}$
\begin{equation*} A= \begin{pmatrix} 1&0&1\\0&1&2\\1&2&1 \end {pmatrix}. \end{equation*}

Solution: exploit the zero in the $(1,2)$-slot of $A$ to see that $e_1,e_2,y=(-1,-2,1)$ is a diagonalising basis and so gives us a diagonal matrix representing $B$ with $Q(e_1)=Q(e_2)=1>0$ and $Q(y)=-4<0$ along the diagonal. So the signature is $(2,1)$.

Here are two alternative techniques:
- (1) Orthogonal diagonalisation yields a diagonal matrix representing $B$ with the eigenvalues of $A$ down the diagonal so we just count how many positive and negative eigenvalues there are.
  
  In fact, $A$ has eigenvalues $1$ and $1\pm \sqrt {5}$. Since $\sqrt {5}>2$, $1-\sqrt {5}<0$ and we again conclude that the signature is $(2,1)$.
  
  Danger: this method needed us to solve a cubic equation which is already difficult. For an $n\times n$ $A$ with $n\geq 5$, this could be impossible!
- (2) Finally, we could try and write $Q$ as a linear combination of linearly independent squares and then count the number of positive and negative coefficients. In fact,
  $\seteqnumber{0}{6.}{1}$
  \begin{align*} Q(x)&=x_1^2+x_2^2+x_3^2+2x_1x_3+4x_2x_3\\ &=(x_1+x_3)^2+x_2^2+4x_2x_3=(x_1+x_3)^2+(x_2+2x_3)^2-4x_3^2. \end{align*} But now we need to check that $x_1+x_3, x_2+2x_3,x_3$ are linearly independent linear functionals on $\R ^3$. Here Corollary 5.7 comes to the rescue and says we only need show that $(\ker x_1+x_3)\cap (\ker x_2+2x_3)\cap (\ker x_3)=\set {0}$. But $x_3=0=x_1+x_3=x_2+2x_3$ rapidly implies that each $x_i=0$ and we are done. The coefficients of these squares are $1,1,-4$ and so, once more, we get that the signature is $(2,1)$.