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I am having a tough time understanding adjoint of a linear map. Consider a linear map between two vector spaces $\, f:V\rightarrow W,$ let us denote $f^*$ to denote its adjoint.

Can some body clarify this discrepancy?

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    $\begingroup$ Without looking at the links: the second one probably considers only Hilbert spaces. We have a concept of a Hilbert space adjoint between the spaces themselves, and the more general concept of an adjoint mapping the duals. The Hilbert space adjoint corresponds to the general adjoint under the Riesz anti-isomorphism between a Hilbert space and its dual. $\endgroup$ – Daniel Fischer May 3 '16 at 14:54
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The adjoint of a linear map $T: \Bbb V \to \Bbb W$ between two vector spaces is given by the definition in the first source: It is the map $T^* : \Bbb W^* \to \Bbb V^*$ defined by $$(T^*(\phi))(v) := \phi(T(v))$$ for all $\phi \in \Bbb W^*$ and $v \in \Bbb V$.

In the second source, $\Bbb V$ and $\Bbb W$ are inner product spaces. An inner product $\langle \,\cdot\, , \,\cdot\, \rangle$ on a vector space $\Bbb U$ defines an isomorphism $\Phi : \Bbb U \stackrel{\cong}{\to} \Bbb U^*$ by $$(\Phi(u))(u') := \langle u, u' \rangle .$$ Thus, for any linear map $T: \Bbb V \to \Bbb W$ between inner product spaces, we can identify $\Bbb W^*$ with $\Bbb W$ and $\Bbb V^*$ with $\Bbb V$, and hence $T^*$ with a map $\Bbb W \to \Bbb V$. Unwinding the definitions shows that this map satisfies the identity $$\langle w, T v \rangle = \langle T^* w, v \rangle$$ in the second definition.

It is an instructive exercise to write out all of these objects in terms of their matrix representations with respect to some bases of $\Bbb V, \Bbb W$ (of course, this only makes sense in the case that the vector spaces are finite-dimensional but even in the infinite-dimensional case it is a useful mnemonic). In particular, if $\Bbb V, \Bbb W$ are finite-dimensional real vector spaces and we choose orthonormal bases of both spaces, one can show that the matrix representations of $T^*$ and $T$ are related by $[T^*] = [T]^{\top}$.

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  • $\begingroup$ In $[T^*] = {}^t[T]$ does ${}^t[T]$ transpose of the matrix representation of $T$? $\endgroup$ – Coniferous May 3 '16 at 18:54
  • $\begingroup$ Yes; usually I prefer the notation ${}^T$, but I thought it would look peculiar here given the use of $T$ for the transformation. $\endgroup$ – Travis Willse May 3 '16 at 19:11
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If the vector spaces V and W have respective nondegenerate bilinear forms $B_V$ and $B_W$, a concept closely related to the transpose – the adjoint – may be defined:

If $f : V → W$ is a linear map between vector spaces $V$ and $W$, we define $g$ as the adjoint of f if $g : W → V$ satisfies :

$ B_V(υ,g(w))=B_W(f(υ),w)$ $ \forall υ \in V, w\in W $.

These bilinear forms define an isomorphism between $V $and $V^∗$, and between W and $W^∗$, resulting in an isomorphism between the transpose and adjoint of $f$. The matrix of the adjoint of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms. In this context, many authors use the term transpose to refer to the adjoint as defined here.

The adjoint allows us to consider whether $g : W → V$ is equal to $f^{−1} : W → V$. In particular, this allows the orthogonal group over a vector space $V$ with a quadratic form to be defined without reference to matrices (nor the components thereof) as the set of all linear maps $V → V$ for which the adjoint equals the inverse

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