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Suppose that $V$ and $W$ are 2 finite dimensional vectors spaces and $T$ is a linear transformation such that $T : V \rightarrow W$. Then $T \rightarrow T^t$ can be seen as an isomorphism of $L(V,W)$ into $L(W^\ast,V^\ast)$.

Now suppose that $V$ and $W$ are infinite dimensional vector spaces, can a transpose exist and is there an isomorphism?

darkgbm
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    AFAIK, the adjoint of a linear map between infinite-dimensional spaces is the infinite-dimensional version of a conjugate transpose matrix between finite-dimensional linear spaces. Note that both satisfy $<Ax,y>=<x,A^ty>$. – user99680 Nov 28 '13 at 06:50
  • What does $<Ax,y>=<x,A^ty>$ mean? I have not learnt adjoint yet... – darkgbm Nov 28 '13 at 06:54
  • $<Ax,y>$ just means the inner-product of $Ax$ and $y$. This is true when you take the conjugate transpose of the matrix $A$, and $x,y$ are just any vectors in your space. – user99680 Nov 28 '13 at 06:57
  • What does this fact say about whether there is an isomorphism or not? – darkgbm Nov 28 '13 at 07:02
  • I think it says you have to examine what is actually meant by the transpose of a linear operator (rather than a matrix of said operator). – R R Nov 28 '13 at 07:10
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    Note that the adjoint of a map does not always exist; they do for bounded operators in Hilbert space (using Riesz representation theorem), but they often do not exist. – user99680 Nov 28 '13 at 07:10
  • So, at the end of the day, there can be no isomorphism (outside of the case of bounded operators on Hilbert space), since not every map has an adjoint. – user99680 Nov 28 '13 at 07:17

1 Answers1

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In the infinite dimensional case, the transpose of a matrix becomes the adjoint map of a linear map $T$ . But, unlike the case for finite-dimensional spaces, the adjoint of a map often does not exist. One of the nicest cases where adjoints always exist is the case of bounded operators in Hilbert space, for which, using the Riesz representation theorem, the adjoint of a map always exists. So, outside of this case, the possible isomorphism fails because a map may not have an adjoint defined.

EDIT: You may be interested in this link: Adjoint of a linear transformation in an infinite dimension inner product space

Which says much of what I said, but also includes detailed examples.

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user99680
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