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I am currently trying to prove that the matrix representation of a dual map ($S^*$) is the transpose of the matrix representation of the linear transformation $S$. I have seen a proof online but it was a little difficult for me to follow.

So far, I have the following:

Let $V$ and $W$ be vector spaces with ordered bases $\beta = \{\mathbb{v}_1, \mathbb{v}_2, \dots, \mathbb{v}_n\}$ and $\gamma = \{\mathbb{w}_1, \mathbb{w}_2, \dots, \mathbb{w}_m\}$ respectively. Let the respective dual space bases be denoted $\beta^* = \{\phi_1, \phi_2, \dots, \phi_n\}$ and $\gamma^* = \{\varphi_1, \varphi_2, \dots, \varphi_m\}$. Additionally, consider $S \in \mathcal{L}(V,W)$. I want to show: $$[S]_\beta^\gamma = [S^*]_{\gamma^*}^{\beta^*}$$ where S is represented by $[S]_\beta^\gamma = (a_{ij})$.

At this point, I am lost as where to go. I think the next step might be to work with the matrix representation $[S^*]_{\gamma^*}^{\beta^*}$?

Any hints as to what approach I should take would be appreciated.

CoffeeDonut
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  • I answered this question here. Is that proof also difficult to follow? – Ben Grossmann Apr 26 '17 at 14:00
  • I am particularly confused about the appearance of $\delta$. Not sure where that comes from but maybe I just need more coffee. – CoffeeDonut Apr 26 '17 at 14:06
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    Or donuts, perhaps. In any case: $\delta_{ij}$ (or $\delta^i_j$, as I wrote it) is the "Dirac delta", defined by $$ \delta_{ij} = \begin{cases} 1 & i=j\ 0 & i \neq j \end{cases} $$ If it helps, you could just skip the step with the $\delta$ and simply note that when we write that first line, we're really adding a bunch of $0$-terms. – Ben Grossmann Apr 26 '17 at 14:44
  • Oh okay, I am aware of that but had not seen the $\delta$ notation before. I'll take another look at your proof, thanks! – CoffeeDonut Apr 27 '17 at 00:05
  • @amd "Let the respective dual bases be denoted..." – Ben Grossmann Apr 27 '17 at 02:24

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