MA22020: Advanced linear algebra

Chapter 5 Symmetric bilinear forms and quadratic forms

We give describe a generalisation of real inner products to vectors spaces \(V\) over an arbitrary field \(\F \) and use this to study the simplest non-linear functions on \(V\).

5.1 Bilinear forms and matrices

Definition. Let \(V\) be a vector space over a field \(\F \). A map \(B:V\times V\to \F \) is bilinear if it is linear in each slot separately:
\(\seteqnumber{0}{5.}{0}\)
\begin{align*} B(\lambda v_1+v_2,v)&=\lambda B(v_1,v)+B(v_2,v)\\ B(v,\lambda v_1+v_2)&=\lambda B(v,v_1)+B(v,v_2), \end{align*} for all \(v,v_1,v_2\in V\), \(v,v_1,v_2\in V\) and \(\lambda \in \F \).

A bilinear map \(V\times V\to \F \) is called a bilinear form on \(V\).

Remark. A bilinear form \(B:V\times V\to \F \) has \(B(v,0)=B(0,v)=0\), for all \(v\in V\). Indeed,
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} B(v,0)=B(v,0+0)=B(v,0)+B(v,0) \end{equation*}

and similarly for \(B(0,v)\).

Examples.
- (1) Any real inner product is a bilinear form (what goes wrong for complex inner products?).
- (2) Let \(A\in M_{n}(\F )\) and define a bilinear form \(B_A:\F ^n\times \F ^n\to \F \) by
  \(\seteqnumber{0}{5.}{0}\)
  \begin{equation*} B_A(x,y)=\bx ^TA\by . \end{equation*}
  
  This gives us a new use for matrices.

There is a converse to this last example:

Definition. Let \(V\) be a vector space over \(\F \) with basis \(\cB =\lst {v}1n\) and let \(B:V\times V\to \F \) be a bilinear form. The matrix of \(B\) with respect to \(\cB \) is \(A\in M_{n}(\F )\) given by
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} A_{ij}=B(v_i,v_j), \end{equation*}

for \(\bw 1{i,j}n\).

The matrix \(A\) along with \(\cB \) tells the whole story:

Proposition 5.1. Let \(B:V\times V\to \F \) be a bilinear form with matrix \(A\) with respect to \(\cB =\lst {v}1n\). Then \(B\) is completely determined by \(A\): if \(v=\sum _{i=1}^nx_iv_i\) and \(w=\sum _{j=1}^ny_jv_j\) then
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} B(v,w)=\sum _{i,j=1}^nx_iy_jA_{ij}=\bx ^TA\by . \end{equation*}

Proof. We simply expand out using the bilinearity of \(B\):
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} B(v,w)=\sum _{i,j=1}^nx_iy_jB(v_i,v_j)=\sum _{i,j=1}^nx_iy_jA_{ij}. \end{equation*}

□

Remark. When \(V=\F ^n\) and \(\cB :\lst {e}1n\) is the standard basis, this tells us that any bilinear form on \(V\) is \(B_{A}\) where \(A_{ij}=B(e_i,e_j)\).

How does \(A\) change when we change basis of \(V\)?

Proposition 5.2. Let \(B:V\times V\to \F \) be a bilinear form with matrices \(A\) and \(A'\) with respect to bases \(\cB :\lst {v}1n\) and \(\cB ':\lst {v'}1n\) of \(V\). Then
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} A'=P^TAP \end{equation*}

where \(P\) is the change of basis matrix¹from \(\cB \) to \(\cB '\): thus \(v'_j=\sum _{i=1}^nP_{ij}v_i\), for \(\bw 1jn\).

¹ Algebra 1B, Definition 1.6.1.

Proof. Using the bilinearity to expand things out, we compute:
\(\seteqnumber{0}{5.}{0}\)
\begin{multline*} A'_{ij}=B(v'_i,v'_j)=B(\sum _kP_{ki}v_k,\sum _hP_{hj}v_{h})\\ =\sum _{k,h}P_{ki}B(v_k,v_h)P_{hj}=\sum _{k,h}(P^T)_{ik}A_{kh}P_{hj}=(P^TAP)_{ij}. \end{multline*} □

This prompts:

Definition. We say that matrices \(A,B\in M_n(\F )\) are congruent if there is \(P\in \GL (n,\F )\) such that
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} B=P^TAP. \end{equation*}