| Version 44 |
Version 43 |
| \paragraph{Definition.} |
\paragraph{Definition.} |
| Let $U,V,W$ be vector spaces over a field $K$. A \emph{bilinear map} |
Let $U,V,W$ be vector spaces over a field $K$. A \emph{bilinear map} |
| is a function $B: U |
is a function $B: U |
| \times V \to W$ such that |
\times V \to W$ such that |
| \begin{enumerate} |
\begin{enumerate} |
| \item the map $x \mapsto B(x,y)$ from $U$ to $W$ is linear for each $y |
\item the map $x \mapsto B(x,y)$ from $U$ to $W$ is linear for each $y |
| \in V$ |
\in V$ |
| \item the map $y \mapsto B(x,y)$ from $V$ to $W$ is linear for each $x |
\item the map $y \mapsto B(x,y)$ from $V$ to $W$ is linear for each $x |
| \in U$. |
\in U$. |
| \end{enumerate} |
\end{enumerate} |
| That is, $B$ is \emph{bilinear} if it is linear in each parameter, |
That is, $B$ is \emph{bilinear} if it is linear in each parameter, |
| taken separately. |
taken separately. |
|
|
| \paragraph{Bilinear forms.} |
\paragraph{Bilinear forms.} |
| A \emph{bilinear form} is a bilinear map $B:V\times V\to K$. A |
A \emph{bilinear form} is a bilinear map $B:V\times V\to K$. A |
| $W$-valued bilinear form is a bilinear map $B: V\times V\to W$. One |
$W$-valued bilinear form is a bilinear map $B: V\times V\to W$. One |
| often encounters bilinear forms with additional assumptions. A |
often encounters bilinear forms with additional assumptions. A |
| bilinear form is called |
bilinear form is called |
| \begin{itemize} |
\begin{itemize} |
| \item \emph{symmetric} if $B(x,y) = B(y,x)$, $x,y \in V$; |
\item \emph{symmetric} if $B(x,y) = B(y,x)$, $x,y \in V$; |
| \item \emph{skew-symmetric} if $B(x,y) = -B(y,x)$, $x,y \in V$; |
\item \emph{skew-symmetric} if $B(x,y) = -B(y,x)$, $x,y \in V$; |
| \item \emph{alternating} if $B(x,x) = 0$, $x \in V$. |
\item \emph{alternating} if $B(x,x) = 0$, $x \in V$. |
| \end{itemize} |
\end{itemize} |
| By expanding $B(x+y,x+y) = 0$, we can show alternating implies |
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. Further if $K$ is not of characteristic $2$, then |
| skew-symmetric implies alternating. |
skew-symmetric implies alternating. |
|
|
| \paragraph{Left and Right Maps.} |
\paragraph{Left and Right Maps.} |
| Let $B:U\times V\to W$ be a bilinear map. We may identify $B$ with the |
Let $B:U\times V\to W$ be a bilinear map. We may identify $B$ with the |
| linear map $B_\otimes: U\otimes V\to W$ (see tensor product). We may |
linear map $B_\otimes: U\otimes V\to W$ (see tensor product). We may |
| also identify $B$ with the linear maps |
also identify $B$ with the linear maps |
| \begin{align*} |
\begin{align*} |
| & B_L:U\to L(V,W),\qquad B_L(x)(y) = B(x,y),\; x\in U,\; y\in V;\\ |
& B_L:U\to L(V,W),\qquad B_L(x)(y) = B(x,y),\; x\in U,\; y\in V;\\ |
| & B_R:V\to L(U,W),\qquad B_R(y)(x) = B(x,y),\; x\in U,\; y\in V. |
& B_R:V\to L(U,W),\qquad B_R(y)(x) = B(x,y),\; x\in U,\; y\in V. |
| \end{align*} |
\end{align*} |
| called the left and right map, respectively. |
called the left and right map, respectively. |
|
|
| Next, suppose that $B:V\times V\to K$ is a bilinear form. Then both |
Next, suppose that $B:V\times V\to K$ is a bilinear form. Then both |
| $B_L$ and $B_R$ are linear maps from $V$ to $V^*$, the dual vector |
$B_L$ and $B_R$ are linear maps from $V$ to $V^*$, the dual vector |
| space of $V$. We can therefore say that $B$ is symmetric if and only |
space of $V$. We can therefore say that $B$ is symmetric if and only |
| if $B_L = B_R$ and that $B$ is anti-symmetric if and only if |
if $B_L = B_R$ and that $B$ is anti-symmetric if and only if |
| $B_L=-B_R$. If $V$ is finite-dimensional, we can identify $V$ and |
$B_L=-B_R$. If $V$ is finite-dimensional, we can identify $V$ and |
| $V^{**}$, and assert that $B_L=(B_R)^*$; the left and right maps are, in |
$V^{**}$, and assert that $B_L=(B_R)^*$; the left and right maps are, in |
| fact, dual homomorphisms. |
fact, dual homomorphisms. |
|
|
|
|
| \paragraph{Rank.} |
\paragraph{Rank.} |
| Let $B:U\times V\to W$ be a bilinear map, and suppose that $U,V$ are |
Let $B:U\times V\to W$ be a bilinear map, and suppose that $U,V$ are |
| finite dimensional. One can show that $\rank B_L = \rank B_R$. We call |
finite dimensional. One can show that $\rank B_L = \rank B_R$. We call |
| this integer $\rank B$, the \emph{\PMlinkescapetext{rank}} of $B$. |
this integer $\rank B$, the \emph{\PMlinkescapetext{rank}} of $B$. |
| Applying the rank-nullity theorem to both the left and right maps |
Applying the rank-nullity theorem to both the left and right maps |
| gives the following results: |
gives the following results: |
| \begin{align*} |
\begin{align*} |
| \dim U &= \dim \ker {B_L} + \rank B\\ |
\dim U &= \dim \ker {B_L} + \rank B\\ |
| \dim V &= \dim \ker {B_R} + \rank B |
\dim V &= \dim \ker {B_R} + \rank B |
| \end{align*} |
\end{align*} |
| We say that $B$ is |
We say that $B$ is |
| \emph{non-degenerate} if both the left and right map are |
\emph{non-degenerate} if both the left and right map are |
| non-degenerate. Note that in order for $B$ to be non-degenerate it is |
non-degenerate. Note that in order for $B$ to be non-degenerate it is |
| necessary that $\dim U = \dim V$. If this holds, then $B$ is |
necessary that $\dim U = \dim V$. If this holds, then $B$ is |
| non-degenerate if and only if $\rank B$ is equal to $\dim U, \dim V$. |
non-degenerate if and only if $\rank B$ is equal to $\dim U, \dim V$. |
|
|
| \paragraph{Orthogonal complements.} |
\paragraph{Orthogonal complements.} |
| Let $B:V\times V\to K$ be a bilinear form, and let $S \subset V$ be a |
Let $B:V\times V\to K$ be a bilinear form, and let $S \subset V$ be a |
| subspace. The left and right orthogonal complements of $S$ are |
subspace. The left and right orthogonal complements of $S$ are |
| subspaces ${^\perp}S, S^\perp \subset V$ defined as follows: |
subspaces ${^\perp}S, S^\perp \subset V$ defined as follows: |
| \begin{align*} |
\begin{align*} |
| &{}^\perp S = \{ u\in V \mid B(u,v) = 0 \; \text{for all } v \in S \},\\ |
&{}^\perp S = \{ u\in V \mid B(u,v) = 0 \; \text{for all } v \in S \},\\ |
| &S^\perp = \{ v\in V \mid B(u,v) = 0 \; \text{for all } u \in S \} . |
&S^\perp = \{ v\in V \mid B(u,v) = 0 \; \text{for all } u \in S \} . |
| \end{align*} |
\end{align*} |
| We may also realize $S^\perp$ by considering the linear map $B_{R}': |
We may also realize $S^\perp$ by considering the linear map $B_{R}': |
| V\to S^*$ obtained as the composition of $B_R: |
V\to S^*$ obtained as the composition of $B_R: |
| V\to V^*$ and the dual homomorphism $V^*\to S^*$. |
V\to V^*$ and the dual homomorphism $V^*\to S^*$. |
| Indeed, $S^\perp = \ker B^\prime_R$. An analogous |
Indeed, $S^\perp = \ker B^\prime_R$. An analogous |
| statement can be made for ${}^\perp S$. |
statement can be made for ${}^\perp S$. |
|
|
| Next, suppose that $B$ is non-degenerate. |
Next, suppose that $B$ is non-degenerate. |
| By the rank-nullity |
By the rank-nullity |
| theorem we have that |
theorem we have that |
| \begin{align*} |
\begin{align*} |
| \dim V &= \dim S + \dim S^\perp\\ |
\dim V &= \dim S + \dim S^\perp\\ |
| &= \dim S + \dim {}^\perp S. |
&= \dim S + \dim {}^\perp S. |
| \end{align*} |
\end{align*} |
| Therefore, if $B$ is non-degenerate, then |
Therefore, if $B$ is non-degenerate, then |
| \[\dim S^\perp = \dim {}^\perp S. \] |
\[\dim S^\perp = \dim {}^\perp S. \] |
| Indeed, more can be said if $B$ is either symmetric or |
Indeed, more can be said if $B$ is either symmetric or |
| skew-symmetric. In this case, we actually have |
skew-symmetric. In this case, we actually have |
| \[{}^\perp S = S^\perp.\] |
\[{}^\perp S = S^\perp.\] |
|
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|
|
| We say that $S\subset V$ is a non-degenerate subspace relative to $B$ |
We say that $S\subset V$ is a non-degenerate subspace relative to $B$ |
| if the restriction of $B$ to $S\times S$ is non-degenerate. Thus, $S$ |
if the restriction of $B$ to $S\times S$ is non-degenerate. Thus, $S$ |
| is a non-degenerate subspace if and only if $S \cap S^\perp = \{0\}$, |
is a non-degenerate subspace if and only if $S \cap S^\perp = \{0\}$, |
| and also $S\cap {}^\perp S = \{ 0\}$. Hence, if $B$ is non-degenerate |
and also $S\cap {}^\perp S = \{ 0\}$. Hence, if $B$ is non-degenerate |
| and if $S$ is a non-degenerate |
and if $S$ is a non-degenerate |
| subspace, we have |
subspace, we have |
| \[ V = S \oplus S^\perp = S\oplus {}^\perp S.\] Finally, note that if |
\[ V = S \oplus S^\perp = S\oplus {}^\perp S.\] Finally, note that if |
| $B$ is positive-definite, then $B$ is necessarily non-degenerate and |
$B$ is positive-definite, then $B$ is necessarily non-degenerate and |
| that every subspace is non-degenerate. In this way we arrive at the |
that every subspace is non-degenerate. In this way we arrive at the |
| following well-known result: if $V$ is positive-definite inner product |
following well-known result: if $V$ is positive-definite inner product |
| space, then |
space, then |
| \[V = S \oplus S^\perp \] for every subspace $S\subset V$. |
\[V = S \oplus S^\perp \] for every subspace $S\subset V$. |
|
|
| \paragraph{Adjoints.} |
\paragraph{Adjoints.} |
| Let $B: V \times V \rightarrow K$ be a non-degenerate bilinear |
Let $B: V \times V \rightarrow K$ be a non-degenerate bilinear |
| form, and let $T \in L(V,V)$ be a linear endomorphism. We define the |
form, and let $T \in L(V,V)$ be a linear endomorphism. We define the |
| right adjoint $T^\star \in |
right adjoint $T^\star \in |
| L(V,V)$ to be the unique linear map such that |
L(V,V)$ to be the unique linear map such that |
| \[ B(Tu, v) = B(u, T^\star v) ,\quad u,v \in V.\] |
\[ B(Tu, v) = B(u, T^\star v) ,\quad u,v \in V.\] |
| Letting $T^\ast : V^\ast \to V^\ast$ denote the dual homomorphism, |
Letting $T^\ast : V^\ast \to V^\ast$ denote the dual homomorphism, |
| we also have |
we also have |
| \[ T^\star = {B_R}^{-1} \circ T^\ast \circ B_R.\] |
\[ T^\star = {B_R}^{-1} \circ T^\ast \circ B_R.\] |
| Similarly, we define the left adjoint ${}^\star T\in L(V,V)$ by |
Similarly, we define the left adjoint ${}^\star T\in L(V,V)$ by |
| \[ {}^\star T = {B_L}^{-1} \circ T^\ast \circ B_L.\] We then have |
\[ {}^\star T = {B_L}^{-1} \circ T^\ast \circ B_L.\] We then have |
| \[ B(u, Tv) = B({}^\star T u, v) ,\quad u,v \in V.\] |
\[ B(u, Tv) = B({}^\star T u, v) ,\quad u,v \in V.\] |
| If $B$ is either symmetric or skew-symmetric, then ${}^\star T = |
If $B$ is either symmetric or skew-symmetric, then ${}^\star T = |
| T^\star$, and we simply use $T^\star$ to refer to the adjoint homomorphism. |
T^\star$, and we simply use $T^\star$ to refer to the adjoint homomorphism. |
|
|
| \paragraph{Additional remarks.} |
\paragraph{Additional remarks.} |
|
|
| \begin{enumerate} |
\begin{enumerate} |
| \item |
\item |
| if $B$ is a symmetric, non-degenerate bilinear form, |
if $B$ is a symmetric, non-degenerate bilinear form, |
| then the adjoint operation is represented, relative to an orthogonal |
then the adjoint operation is represented, relative to an orthogonal |
| basis (if one exists), by the matrix transpose. |
basis (if one exists), by the matrix transpose. |
|
|
| \item If $B$ is a symmetric, non-degenerate bilinear form |
\item If $B$ is a symmetric, non-degenerate bilinear form |
| then $T\in L(V,V)$ is then said to be a \emph{normal operator} (with |
then $T\in L(V,V)$ is then said to be a \emph{normal operator} (with |
| respect to $B$) if $T$ commutes with its adjoint $T^\star$. |
respect to $B$) if $T$ commutes with its adjoint $T^\star$. |
|
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|
| \item An $n \times m$ matrix may be regarded as a bilinear form over |
\item An $n \times m$ matrix may be regarded as a bilinear form over |
| $K^n\times K^m$. Two such matrices, $B$ and $C$, are said to be |
$K^n\times K^m$. Two such matrices, $B$ and $C$, are said to be |
| congruent if there exists an invertible $P$ such that $B = P^{T}CP$. |
congruent if there exists an invertible $P$ such that $B = P^{T}CP$. |
|
|
| \item The identity matrix, $I_n$ on $\Rset^n\times \Rset^n$ gives the standard |
\item The identity matrix, $I_n$ on $\Rset^n\times \Rset^n$ gives the standard |
| Euclidean inner product on $\Rset^n$. |
Euclidean inner product on $\Rset^n$. |
| \end{enumerate} |
\end{enumerate} |
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