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I am extremely new to the concept of dual space and the notation $\mathcal{L}(V,W)$ which denotes the set of all linear transformations from $V\to W$. I read that the dual space $V^*=\mathcal{L}(V,\Bbb{K})$ is isomorphic to $V$, where $V$ is some vector space defined over $\Bbb{K}$. The “proof” (or perhaps intuition) that was given to me was that $\dim\mathcal{L}(V,\Bbb{K})=\dim(V)\dim(\Bbb{K})=\dim(V)$.

Two questions:

  1. Rigorously, and I apologise if this is naïve but as I say I have never seen these concepts until approximately five minutes ago, why is $\dim\mathcal{L}(A,B)=\dim(A)\dim(B)$?

  2. How is it a complete proof to just show that the dimensions of $V^*$ and $V$ are the same? Surely some more work needs to be done to show isomorphisms? For context, I have never actually seen a proof of isomorphism and do not know what proving an isomorphism in an abstract context like this entails.

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It's important that we're talking about finite-dimensional vector spaces here. If $\dim(A)=n$ and $\dim(B)=m$, then there's a $1$-$1$ correspondence between linear transformations $f:A \to B$ and $m \times n$ matrices. That's because any such transformation is determined by its values on a basis for $A$, and those values in turn are determined by a column vector with $m$ components. The dimension of the space of $m \times n$ matrices is $mn$.

It's also a theorem that any two finite-dimensional vector spaces over the same field with the same dimension are isomorphic to one another. Choose a basis for each and map element $k$ of the basis for $A$ to element $k$ of the basis for $B$. That results in a linear map that is onto a basis of $B$, and therefore is onto $B$, and that has trivial kernel.

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  • $\begingroup$ Does this notion of a “vector” being a linear combination of the basis hold even when the “vector” space in question is something more abstract, such as a function space? $\endgroup$
    – FShrike
    Aug 10 at 22:46
  • $\begingroup$ Yes, but such spaces are pretty likely to be infinite-dimensional, which adds complications. The basic results still hold but you start needing things like the Axiom of Choice to prove them. In that context, it's important to note that a vector is a linear combination of a finite number of elements from your basis. $\endgroup$ Aug 10 at 22:47
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    $\begingroup$ Thank you for the clarification $\endgroup$
    – FShrike
    Aug 10 at 22:54

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