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Let $f \in \mathcal{L}(V, W)$. Moreover let's suppose $(e_1, ..., e_n)$ is a basis of $V$ and $f(e_i) = v_i$ where the $v_i$ aren't distinct (so there is at least $i \ne j$ such that $v_i = v_j$) so that $f$ is not one-to-one.

Then what I don't get is that the adjoint $f^* : W^* \to V^*$ is defined as : $f(v_i^*) = e_i^*$, but it doens't mean anything when $f$ is not one-to-one right ?

So why for example here they are talking about the linear transformation in the dual associated with $f$ which is $f$ but in the basis $B^*$ of $W^*$ and $C^*$ of $W^*$ ?

Thank you !

José Carlos Santos
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dghkgfzyukz
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  • What do you mean that it doesn't mean anything? It looks fine to me and does not need the function to be injective to make sense. – Tobias Kildetoft Feb 06 '19 at 09:10
  • The adjoint is typically defined implicitly by $\langle f(v), w \rangle = \langle v, f^(w) \rangle$. Even when $f$ is not one-to-one, there is a unique linear transformation $f^$ that makes the above true. – Theo Bendit Feb 06 '19 at 09:11
  • @TobiasKildetoft If $f(e_1) = v_1$ and $f(e_2) = v_1$ then how can I defined $f^(v_1^)$ ? I mean is $f^(v_1^) = e_1^$ or $f^(v_1^) =e_2^$ ? – dghkgfzyukz Feb 06 '19 at 09:12
  • @TheoBendit Thank you but I am aware of this definition, the thing is that I don't understand how it relates to the other one. – dghkgfzyukz Feb 06 '19 at 09:13
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    @dghkgfzyukz Apologies, I missed that this was in a general finite dimensional linear space, not an inner product space. The way it connects to the inner product version is via the Riesz Representation Theorem. Each $w \in W$ can be thought of as an element of $W^$ by identifying it with the map $x \mapsto \langle x, w \rangle$. Then $f^$ at $w \in W^$ is a linear map $x \mapsto \langle f(x), w \rangle$, according to Jose's (correct) definition. But, this linear map can be uniquely expressed in the form $\langle x, v^ \rangle$ for some $v^$. We identify this $v^$ with $f^*(w)$. – Theo Bendit Feb 06 '19 at 09:33

1 Answers1

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That is not how the adjoint map $f^*$ is defined. The definition is$$(\forall\alpha\in W^*):f^*(\alpha)=\alpha\circ f.$$

When we have a subset $\{f_1,\ldots,f_n\}$ of a vector space $V$, we define $f_1^*,\ldots,f_n^*\in V^*$ by$$f_i^*(f_j)=\begin{cases}1&\text{ if }i=j\\0&\text{ otherwise,}\end{cases}$$but this only makes sense if $\{f_1,\ldots,f_n\}$ is a basis. That's not the case with your $v_i$'s.

José Carlos Santos
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