Example 4.5-3 in Kryeszig's Functional Analysis: What is the relation between the matrix of a linear operator and that of its adjoint?

Question

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 $$f_r \left( e_s \right) = \begin{cases} 1 & \mbox{ if } r = s; \\ 0 & \mbox{ if } r \neq s.\end{cases} $$

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Answer 1

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 $A$.

Answer 2

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.