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Let $X$ and $Y$ be finite-dimensional normed spaces, either both real or both complex, and let $T \colon X \to Y$ be a linear operator. (Then $T$ is bounded since its domain is finite-dimensional). Let $X^\prime$ denote the dual space of $X$ (i.e. the normed space of all the bounded linear functionals with domain $X$); since $X$ is finite-dimensional, $X^\prime$ consists of all the linear functionals with domain $X$. And, the same applies to $Y^\prime$.

Then the adjoint operator $T^\times \colon Y^\prime \to X^\prime$ of $T$ is also a bounded linear operator that is defined as follows: Let $g \in Y^\prime$. Then $T^\times (g) = f \in X^\prime$, where $f$ is defined by $$f(x) = g\left( T(x) \right) \ \mbox{ for all } \ x \in X.$$ This operator $T^\times$ is linear and bounded with $$ \Vert T^\times \Vert = \Vert T \Vert.$$

Now let $E = \{ e_1, \ldots, e_n \}$ be an ordered basis for $X$, and let $B = \{ b_1, \ldots, b_m \}$ be an ordered basis for $Y$. Let $A = \left[ \alpha_{ij} \right]_{m \times n}$ be the matrix of $T$ with respect to the ordered bases $E$ and $B$. Let $E^\prime$ and $B^\prime$ denote the dual bases of $E$ and $B$, respectively.

Let $A^\times$ be the matrix of $T^\times$ with respect to the dual bases $B^\prime$ and $E^\prime$.

What is the relation between the matrices $A$ and $A^\times$?

By definition, the dual basis $E^\prime$ of $E$ the set of linear functionals $\{ f_1, \ldots, f_n \}$ with domain $X$ such that, for each $j = 1, \ldots, n$ and for each $k = 1, \ldots, n$,
$$f_j(e_k) = \begin{cases} 1 \ \mbox{ if } \ j = k; \\ 0 \ \mbox{ if } \ j \neq k.\end{cases} $$

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  • $\begingroup$ I think it should be the inverse of the transposed, but I'm not 100% sure. That is, $A^{\times} = (A^{-1})^T$ $\endgroup$ – Riccardo Orlando Jun 19 '16 at 17:12
  • $\begingroup$ There is no reason to require a normed space. Indeed, there is no reason to require field $\mathbb R$ or $\mathbb C$. The question still makes sense. $\endgroup$ – GEdgar Jun 19 '16 at 17:37
  • $\begingroup$ @Riccardo Orlando No need to inverse : transpose only is the right answer. $\endgroup$ – Jean Marie Jun 19 '16 at 19:43
  • $\begingroup$ @JeanMarie oh well. Inverting seemed reasonable since you sort of go backwards when pushing to the duals... That is, $T^{\times}$ goes from $Y'$ to $X'$ $\endgroup$ – Riccardo Orlando Jun 19 '16 at 20:22
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We have that $T^{\times}(b_j^{\times})(e_i)=b_j^{\times}(T(e_i))=b_j^{\times}(\sum\limits_k a_{k,i}b_k )=a_{j,i}.$

Therefore, $$T^{\times}(b_j^{\times})=\sum_{i} a_{j,i}b_i^{\times}.$$

It follows that the matrix representation of the adjoint is the transpose.

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