Is there a simple proof showing that $\dim V^*=\dim V$? Where $V^*$ denotes the dual space of a finite dimensional space $V$?
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You should be able to do it pretty directly by using the dual basis, although note that your statement is only true if $V$ is finite dimensional. – Jakob Hansen Jun 09 '16 at 18:02
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5See also this question. – Dietrich Burde Jun 09 '16 at 18:07
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Depends what you mean by "simple". However, I think this proof is pretty straightforward:
Let $\{v_1, \dots, v_n\}$ be a basis for $V$. Let $\{f_1,\dots,f_n\} \subset V^*$ be defined so that $$ f_i(v_j) = \begin{cases} 1 & i=j\\ 0 & i \neq j \end{cases} $$ (This set is usually called the "dual basis" corresponding to our $v_i$s). Now, show that theses functionals form a basis of $V^*$. It would follow that $V^*$ is $n$-dimensional, as desired.
Note: Recall (or prove) that if $T_1$ and $T_2$ are linear transformations such that $T_1(v_i) = T_2(v_i)$ for each $i$, then $T_1$ and $T_2$ are the same linear transformation.
Ben Grossmann
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Maybe you would please comment on 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 $\mathcal L(V,W) =(\dim V)(\dim W)$ – Mar 30 '17 at 14:55
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1@Andrew that works too, but it assumes his prior result. In fact, we can go in the other direction too: one we have $\dim V^*$, there's a slick approach to the more general result on $L(V,W)$. – Ben Grossmann Mar 30 '17 at 17:55