Consider the definition of dual map, $T'$ in Axler (2015).

If $T\in\mathcal{L}(V,W)$, then the dual map of $T$ is the linear map $T'\in\mathcal{L}(W',V')$ defined by $T'(\varphi)=\varphi\circ T$ for $\varphi\in W'$.

My Question:
(1) I don't understand how the composition $\varphi\circ T$ works. $T$ maps from $V$ to $W$. Then, how does $\varphi\circ T$ gives you a linear map from $V$ to $\mathcal{F}$ since $\varphi$ is an element of $W'$, how does this composition works?

Reference: Axler, Sheldon J. $\textit{Linear Algebra Done Right}$, New York: Springer, 2015.


1 Answer 1


$W'$ is the dual space of $W$: i.e. it is shorthand for $\mathcal{L}(W,\mathcal{F})$.

Thus $\varphi \in W'$ is a map $W \rightarrow \mathcal{F}$, and $\varphi \circ T$ is a map $V \rightarrow W \rightarrow \mathcal{F}$, which means it is an element of $\mathcal{L}(V,\mathcal{F})$ aka $V'$.

Thus $T'$, which takes an element $\varphi \in W'$ and returns $\varphi \circ T \in V'$, is an element of $\mathcal{L}(W',V')$

  • $\begingroup$ Thanks for the response! A quick questions: how does $\varphi\in W'$ imply it is a map from $W$ to $\mathcal{F}$? $\endgroup$ Commented Jul 15, 2019 at 18:58
  • 2
    $\begingroup$ @FrankSwanton That's what $W'$ means. (You should check to see where it was defined in your book.) $\endgroup$ Commented Jul 15, 2019 at 19:07

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