There is this answer which very intuitively answers the question:

Dual space of finite dimensional space V has the same dimension

But I would appreciate help with Axler's proof (3.95) on page 101 of "Linear Algebra Done Right", 3rd ed. There he gives a one liner referring to a prior result: Dimension of $\mathcal L(V,W) =(\dim V)(\dim W)$. I am puzzeled how to apply this as we know $\dim V$, but we are trying to determine, in this case, $\dim W$.

So how do we know $\dim \mathcal L(V,W)$ a priori?


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    $\begingroup$ Showing $\dim L(V,W)=\dim V\dim W$ amounts to the same proof as given at your link, see here. So proving $\dim V'=\dim V$ this way is again a duplicate. $\endgroup$ – Dietrich Burde Mar 30 '17 at 15:09
  • $\begingroup$ @DietrichBurde If I understand your link, it is ostensibly the same as the link I provided. But in the context of your comment and my question, we have: $\dim \mathcal L(V,V')= (\dim V)(\dim V')$. We know $\dim V$, but how do we know what to use for $V'$ on the LHS. Thanks. With regards, $\endgroup$ – user12802 Mar 30 '17 at 15:19
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    $\begingroup$ This is not what we do - see the answer below. We put $W=K^1$, the $1$-dimensional vector space over $K$, and not $W=V'$. $\endgroup$ – Dietrich Burde Mar 30 '17 at 15:20

If $V$ is a $K$-vector space, then $V' = \mathcal{L}(V,K)$ by Definition. As $K$ is a $1$-dimensional $K$-vector space, you get $$\dim V' = \dim \mathcal{L}(V,K) = \dim(V)\dim(K) = \dim(V).$$

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