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Let $X$ and $Y$ be finite-dimensional normed spaces, either both real or both complex, and let $T \colon X \longrightarrow Y$ be a linear operator. (Then by Theorem 2.7-8 in Kreyszig $T$ 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 $X$, again by Theorem 2.7-8 in Kreyszig. And, the same applies to $Y^\prime$.

Then the adjoint operator $T^\times \colon Y^\prime \longrightarrow X^\prime$ 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 $$f(x) = g\big( T(x) \big) \mbox{ for all } x \in X.$$ This operator $T^\times$ is linear and bounded with $$ \left\lVert T^\times \right\rVert = \lVert T \rVert.$$

Now let $E = \left( e_1, \ldots, e_n \right)$ be an ordered basis for $X$, and let $B = \left( b_1, \ldots, b_m \right)$ 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 ordered 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 $X^\prime$ corresponding to the ordered basis $E$ for $X$ is the ordered $n$-tuple $\left( f_1, \ldots, f_n \right)$ of linear functionals with domain $X$ such that, for each $j = 1, \ldots, n$ and for each $k = 1, \ldots, n$, we have $$ f_j \left( e_k \right) = \begin{cases} 1 & \mbox{ if } j = k; \\ 0 & \mbox{ if } j \neq k.\end{cases} $$

And, similarly the dual basis $B^\prime$ of $B$ is the ordered $m$-tuple $\left( g_1, \ldots, g_m \right)$ of linear functionals with domain $Y$ such that, for each $r = 1, \ldots, m$ and for each $s = 1, \ldots, n$, we have $$f_r \left( e_s \right) = \begin{cases} 1 & \mbox{ if } r = s; \\ 0 & \mbox{ if } r \neq s.\end{cases} $$

Here is a Math Stack Exchange of mine which contains the relevant terminology and notation.

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  • $\begingroup$ I think it should be the inverse of the transposed, but I'm not 100% sure. That is, $A^{\times} = (A^{-1})^T$ $\endgroup$ Jun 19, 2016 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, 2016 at 17:37
  • $\begingroup$ @Riccardo Orlando No need to inverse : transpose only is the right answer. $\endgroup$
    – Jean Marie
    Jun 19, 2016 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 $X'$ $\endgroup$ Jun 19, 2016 at 20:22
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    $\begingroup$ @RiccardoOrlando The author has a sight problem. Perhaps it's difficult for him to follow foreign notations... $\endgroup$
    – Compacto
    Feb 25, 2022 at 17:07

2 Answers 2

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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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  • $\begingroup$ I would be grateful if you could amend your answer by including some more detail as well as modifying the notation and terminology consistent with that used in my post (or the reference given herein toward the end). $\endgroup$ Feb 25, 2022 at 15:48
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I think that Aloizio's answer is correct, but still, I'll try to give a more detailed one.

If $A^{\times} = [\alpha_{pq}']$, is the matrix of $T^{\times}$ with respect to the dual bases $B′$ and $E′$, then the number $\alpha_{pq}'$ (element of the $p$-th row, $q$-th column) is equal to the $p$-th coordinate of $T^{\times}(g_q)$ with respect to the basis $E'$. This number is obtained by applying $T^{\times}(g_q)$ to the $p$-th vector of the basis $E$:

$$T^{\times}(g_q)(e_p) = g_q(T(e_p))$$

And this last number is the $q$-th coordinate of $T(e_p)$ with respect to the basis $B$, which is equal to the element of $A$ placed in the $q$-th row, $p$-th column, that is, $\alpha_{qp}$. It follows that $\alpha_{pq}' = \alpha_{qp}$, and so, $A^{\times}$ is the transpose of $A$.

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    $\begingroup$ thank you very much for such a beautifully clear explanation! $\endgroup$ Mar 5, 2022 at 17:23

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