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| Title of object: |
bilinear form |
| Canonical Name: |
BilinearForm |
| Type: |
Definition |
| Created on: |
2002-01-24 16:53:36 |
| Modified on: |
2005-01-06 05:32:46 |
| Classification: |
msc:15A63, msc:11E39, msc:47A07 |
| Defines: |
Sylvester's Law of Inertia |
| Synonyms: |
bilinear form=bilinear
bilinear form=bilinear form |
Preamble:
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Content:
\paragraph{Definition.}
Let $U$ and $V$ be vector spaces over a field $K$. A function $B: U \times V \rightarrow K$ is called a \emph{bilinear map} if
\begin{enumerate}
\item $B(cx_1+x_2,y) = cB(x_1,y) + B(x_2,y)$ , $c\in K$
\item $B(x,cy_1+y_2) = cB(x,y_1) + B(x,y_2)$ , $c\in K$
\end{enumerate}
That is, $B$ is \emph{bilinear} if it is linear in each parameter, taken separately.
\paragraph{Bilinear forms.}
If $U$ = $V$ then $B$ is a \emph{bilinear form}. In this case further assumptions are often made:
\begin{enumerate}
\item $B(x,y) = B(y,x)$, $x,y \in V$ (symmetric)
\item $B(x,y) = -B(y,x)$, $x,y \in V$ (skew-symmetric)
\item $B(x,x) = 0$, $x \in V$ (alternating)
\end{enumerate}
By expanding $B(x+y,x+y) = 0$, we can show alternating implies skew-symmetric. Further if $K$ is not of characteristic $2$, then skew-symmetric implies alternating.
\paragraph{Left and Right Maps.}
We may regard the bilinear map as a left map or right map, as follows:
$B_L \in L(U, V^\ast)$
$B_R \in L(V, U^\ast)$
$B_L(x)(y) = B(x,y)$
$B_R(y)(x) = B(x,y)$
The left is a linear map from $U$ into the dual of $V$. So for example $B$ is skew-symmetric iff $B_L \equiv -B_R$.
\paragraph{Matrix Representation.}
Suppose $U$ and $V$ are finite-dimensional and we have chosen bases, ${{\cal B}_1} = \{e_1, \ldots\}$ and ${{\cal B}_2}=\{f_1, \ldots\}$. Now we define the matrix $C$ with entries $C_{ij} = B(e_i, f_j)$. This will be the matrix associated to $B$ with respect to this basis as follows; If we write $x,y \in V$ as column vectors in terms of the chosen bases, then $B(x,y) = x^T C y$. Further if we choose the corresponding dual bases for $U^\ast$ and $V^\ast$ then $C$ and $C^T$ are the corresponding matrices for $B_R$ and $B_L$, respectively (in the sense of linear maps). Thus we see that a symmetric bilinear form is represented by a symmetric matrix, and similarly for skew-symmetric forms.
Let ${{\cal B}_1^\prime}$ and ${{\cal B}_2^\prime}$ be new bases, and $P$ and $Q$ the corresponding change of basis matrices. Then the new matrix is $C^\prime = P^{T}CQ$.
\paragraph{Rank.}
If $U$ and $V$ are finite dimensional it may be shown that $\operatorname{rank} B_L = \operatorname{rank} B_R$. We call this simply the \emph{rank} of $B$. We say that $B$ is \emph{non-degenerate} if the left and right map are both linear isomorphisms.
Now applying the rank-nullity theorem on both left and right maps gives the following results:
$$\dim U = \dim \ker {B_L} + r$$
$$\dim V = \dim \ker {B_R} + r$$
\paragraph{Orthogonals.}
If $T \subset U$ and $S \subset V$ then we may define the orthogonals $T^\perp \subset V$ and ${^\perp}S \subset U$ as follows:
$$ T^\perp = \{ v \mid B(t,v) = 0 \; \forall t \in T \} $$
$$ {^\perp}S = \{ u \mid B(u,s) = 0 \; \forall s \in S \} $$
The orthogonal of a subspace is itself a subspace. Further if $B$ is a symmetric or skew-symmetric bilinear form, then $^\perp A = A^\perp$, and we may choose the latter notation.
$T$ is a non-degenerate subspace if $T \cap T^\perp = \{0\}$. Similarly when $S \cap {^\perp}S = \{0\}$.
We may also realise $T^\perp$ by considering the restriction $B^\prime = B_{{T \times V}}$. It's clear that $T^\perp = \ker B^\prime_R$. Now if $B$ is non-degenerate (or more generally $T \cap {^\perp}V = \{0\}$) then $\ker B^\prime_L = \{0\}$, and we can use the rank-nullity equations to get $\dim V = \dim T + \dim T^\perp$. Similarly we may show that $\dim U = \dim S + \dim ^\perp S$.
\paragraph{Canonical Representations for Symmetric Forms.}
If $B:V \times V \rightarrow K$ is a symmetric bilinear form over a finite-dimensional vector space, then there is an orthogonal basis such that $B$ is represented by
$$
\bordermatrix{& \cr
& a_{1} & 0 & \ldots & 0\cr
& 0 & a_{2} & \ldots & 0\cr
& \vdots & \vdots & \ddots & \vdots\cr
& 0 & 0 &\ldots & a_{n}\cr
}
$$
Denote the rank of $B$ by $r$.
If $K = \mathbb{R}$ we may choose a basis such that $a_1 = \ldots a_t = 1$, $a_{t+1} = \ldots = a_{t+p} = -1$ and $a_{t+p+j} = 0$, for some integers $p$ and $t$. Further these integers are \emph{invariants} of the bilinear form. This is known as \emph{Sylvester's Law of Inertia}. $B$ is \emph{positive definite} iff $t = n$, $p = 0$. Such a form constitutes a \emph{Real Inner Product Space}.
If $K = \mathbb{C}$ we may go further and choose a basis such that $a_1 = \ldots = a_r = 1$ and $a_{r + j} = 0$.
If $K = F_p$ we may choose a basis such that $a_1 = \ldots = a_{r-1} = 1$, $a_{r+j} = 0$ and $a_r = n$ or $a_r = 1$; where $n$ is the least positive quadratic non-residue.
\paragraph{Adjoint.}
Suppose $B: U \times V \rightarrow K$ is a non-degenerate bilinear map. If $T \in L(U,U)$ then we define the adjoint of $T$, $T^\star \in L(V,V)$ to be the unique linear map such that:
$$ B(Tu, v) = B(u, T^\star v) $$
Let $T^\ast : U^\ast \rightarrow U^\ast$ be the dual endormorphism. Then $T^\star = {B_R}^{-1} \circ T^\ast \circ B_R$.
If $U = V$ and we choose a canonical (orthogonal) basis for $B$, then the adjoint corresponds to the matrix transpose.
$T$ is then said to be a \emph{Normal Operator} (with respect to this bilinear map) if it commutes with its adjoint.
\paragraph{Examples.}
An important example is the non-degenerate bilinear map
$$ B:V \times V^\ast \rightarrow K $$
$$ B(v,f) = f(v)$$
Here the orthogonal is exactly the annihilator. This gives the result that $\dim U + \dim U ^\circ = \dim V$.
An $n \times m$ matrix may be regarded as a bilinear form over $K^n$ and $K^m$. Two matrices, $B$ and $C$, are then said to be congruent if there exists an invertible $P$ such that $B = P^{T}CP$. Note this is different to the usual notion of congruence.
If the matrix is the identity, $I$, then this gives the standard Euclidean inner product on $K^n$.
An inner product space on a vector space is a bilinear form if its field is real, but not if it is complex. In fact, the bilinear form associated with a real inner product space is non-degenerate, and every subspace is non-degenerate. So, as we may intuitively expect, $V = U \oplus U^\perp$ for every subspace $U$. |
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