| Version 8 |
Version 7 |
| \paragraph{Definition (the bilinear case).} Let $U$ be a |
\paragraph{Definition (the bilinear case).} Let $U$ be a |
| finite-dimensional vector space over a field $\kfield$, and $B:U\times |
finite-dimensional vector space over a field $\kfield$, and $B:U\times |
| U\to\kfield$ a symmetric, non-degenerate bilinear mapping, for example |
U\to\kfield$ a symmetric, non-degenerate bilinear mapping, for example |
| a real inner product. For an endomorphism $T:U\rightarrow U$ we |
a real inner product. For an endomorphism $T:U\rightarrow U$ we |
| define the adjoint of $T$ relative to $B$ to be the endomorphism |
define the adjoint of $T$ relative to $B$ to be the endomorphism |
| $T\adj:U\rightarrow U$, characterized by |
$T\adj:U\rightarrow U$, characterized by |
| $$B(u,Tv) = B(T\adj u,v),\quad u,v\in U.$$ |
$$B(u,Tv) = B(T\adj u,v),\quad u,v\in U.$$ |
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| It is convinient to identify $B$ |
It is convinient to identify $B$ |
| with a linear isomorphism $B:U\rightarrow U\dual$ in the sense that |
with a linear isomorphism $B:U\rightarrow U\dual$ in the sense that |
| $$B(u,v) = (Bu)(v),\quad u,v\in U.$$ |
$$B(u,v) = (Bu)(v),\quad u,v\in U.$$ |
| We then have |
We then have |
| $$T\adj = B^{-1} T\dual B.$$ |
$$T\adj = B^{-1} T\dual B.$$ |
| To put it another way, $B$ gives an |
To put it another way, $B$ gives an |
| isomorphism between $U$ and |
isomorphism between $U$ and |
| the dual $U^*$, and the |
the dual $U^*$, and the |
| adjoint $T\adj$ is the endomorphism of $U$ that corresponds to the |
adjoint $T\adj$ is the endomorphism of $U$ that corresponds to the |
| \PMlinkname{dual homomorphism}{DualHomomorphism} |
\PMlinkname{dual homomorphism}{DualHomomorphism} |
| $T\dual:U\dual\rightarrow U\dual$. Here is a commutative diagram to |
$T\dual:U\dual\rightarrow U\dual$. Here is a commutative diagram to |
| illustrate this idea: |
illustrate this idea: |
| $$ |
|
| \xymatrix{% |
\xymatrix{% |
| U \ar[r]^{T^\star} \ar[d]^B & U \ar[d]^{B} \\ |
U \ar[r]^{T^\star} \ar[d]^B & U \ar[d]^{B} \\ |
| \;U^* \ar[r]^{T^*} & \;U^{*} |
\;U^* \ar[r]^{T^*} & \;U^{*} |
| } |
|
| $$ |
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| \paragraph{Relation to the matrix transpose.} Let $\bu_1,\ldots,\bu_n$ |
\paragraph{Relation to the matrix transpose.} Let $\bu_1,\ldots,\bu_n$ |
| be a basis of $U$, and let $M\in |
be a basis of $U$, and let $M\in |
| \Mat_{n,n}(\kfield)$ be the matrix of $T$ relative to this basis, i.e. |
\Mat_{n,n}(\kfield)$ be the matrix of $T$ relative to this basis, i.e. |
| $$\sum_j M^j_{\,i}\, \bu_j = T(\bu_i).$$ |
$$\sum_j M^j_{\,i}\, \bu_j = T(\bu_i).$$ |
| Let $P\in\Mat_{n,n}(\kfield)$ denote the matrix of the inner product |
Let $P\in\Mat_{n,n}(\kfield)$ denote the matrix of the inner product |
| relative to the same basis, i.e. |
relative to the same basis, i.e. |
| $$P_{ij} = B(\bu_i,\bu_j).$$ |
$$P_{ij} = B(\bu_i,\bu_j).$$ |
| Then, the representing matrix of $T\adj$ relative to the same basis |
Then, the representing matrix of $T\adj$ relative to the same basis |
| is given by $ P^{-1} M\supt P.$ Specializing further, suppose that the |
is given by $ P^{-1} M\supt P.$ Specializing further, suppose that the |
| basis in question is orthonormal, i.e. that |
basis in question is orthonormal, i.e. that |
| $$B(\bu_i,\bu_j) = \delta_{ij}.$$ |
$$B(\bu_i,\bu_j) = \delta_{ij}.$$ |
| Then, the matrix of $T\adj$ is |
Then, the matrix of $T\adj$ is |
| simply the transpose $M\supt$. |
with suitably matched domains. Furthermore, the dual of the |
|
identity homomorphism is the identity homomorphism of the dual space. |
| \paragraph{The Hermitian (sesqui-linear) case.} |
Thus, using the language of category theory, the dualizing operation |
| If $T:U\rightarrow U$ is an endomorphism of a unitary space (a complex |
can be characterized as the homomorphism action of the contravariant, |
| vector space equipped with a \PMlinkname{Hermitian inner product}{HermitianForm}). In this setting we can define we define |
dual-space functor. |
| the Hermitian adjoint $T\adj:U\rightarrow U$ by means of the familiar |
\paragraph{Relation to the matrix transpose.} The above properties closely |
| adjointness condition |
mirror the algebraic properties of the matrix transpose operation. |
| $$\langle u,Tv\rangle = \langle T\adj u,v\rangle,\quad u,v\in U.$$ |
Indeed, $T^*$ is sometimes referred to as the transpose of $T$, |
|
because at the level of matrices the dual homomorphism is calculated |
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by taking the transpose. |
| However, the analogous operation at the matrix level is the conjugate |
To be more precise, suppose that $U$ and $V$ are finite-dimensional, |
| transpose. Thus, if $M\in \Mat_{n,n}(\cnums)$ is the matrix of $T$ |
and let $M\in \Mat_{n,m}(\kfield)$ be the matrix of $T$ relative to |
| relative to an orthonormal basis, then $\overline{M\supt}$ is the |
some fixed bases of $U$ and $V$. Then, the dual homomorphism $T\dual$ |
| matrix of $T\adj$ relative to the same basis. |
is represented as the transposed matrix $M\supt\in\Mat_{m,n}(\kfield)$ |
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relative to the corresponding dual bases of $U\dual, V\dual$. |
