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This question already has an answer here:

Suppose $V$ and $W$ are finite-dimensioanl vectors spaces. Suppose $T\in\mathcal{L}(V,W)$ and $T'\in\mathcal{L}(W',V')$ where $T'$ denote the dual map of $T$ as defined in Axler (2015).

$\textbf{My Question}$: Are the dimensions of $W$ and $W'$ the same? If so or not so, what is the basic intuition behind this?

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

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marked as duplicate by cmk, Shailesh, YuiTo Cheng, José Carlos Santos linear-algebra Jul 20 at 6:52

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    $\begingroup$ Yes. For finite dimensional spaces, the dimensions of a space and its dual space coincide. $\endgroup$ – Viktor Glombik Jul 19 at 19:18
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    $\begingroup$ Your question doesn't make sense --- $T$ is a linear map, and linear maps don't have dimensions. $\endgroup$ – John Hughes Jul 19 at 19:21
  • $\begingroup$ You may awant to read about the dual basis. This may give a nice mathematical intuition in these matters. $\endgroup$ – DonAntonio Jul 19 at 19:21
  • $\begingroup$ @JohnHughes I fixed it. Sorry! $\endgroup$ – Frank Swanton Jul 19 at 19:22
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Recall that, the set of all linear transformations from a $n$-dimensional vector space $\textsf V$ to a $m$-dimensional vector space $\textsf W$ is isomorphic to the set of matrices $\textsf{M}_{m\times n}(F)$, which has dimension $mn$. In other words : $$\dim \mathcal{L}(\textsf{V},\textsf{W}) = \dim (\textsf V) \dim (\textsf W)$$ Now, as we can always consider the field $F$ as a vector space over itself, of $1$ dimension, it follows that $$\dim(\textsf{W}^*)=\dim \mathcal{L}(\textsf{W}, F) = \dim (\textsf W) \dim (F) = \dim (\textsf W)$$

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