# Dimensions of finite-dimensional vector space and its dual [duplicate]

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.

• Yes. For finite dimensional spaces, the dimensions of a space and its dual space coincide. Jul 19 '19 at 19:18
• Your question doesn't make sense --- $T$ is a linear map, and linear maps don't have dimensions. Jul 19 '19 at 19:21
• You may awant to read about the dual basis. This may give a nice mathematical intuition in these matters. Jul 19 '19 at 19:21
• @JohnHughes I fixed it. Sorry! Jul 19 '19 at 19:22

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 $$\dim(\textsf{W}^*)=\dim \mathcal{L}(\textsf{W}, F) = \dim (\textsf W) \dim (F) = \dim (\textsf W)$$