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The dual space $V^*$ of a vector space $V$ over $F$ is defined as $V\to F$.

This seems like a weird definition. In my limited experience with linear algebra, I've thought of the dual vector of a vector as the row vector version of that (column) vector.

What is the rationale for the dual space concept?

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    $\begingroup$ Pick a base for $V$. What happens when you multiply $v^tw$ as matrices? Well the same thing as applying $v^*$ on $w$, where $v^*$ is the image of $v$ under isomorphism $V\stackrel\sim\to V^*$ that sends base of $V$ to its dual base. $\endgroup$
    – Ennar
    Commented Apr 29, 2018 at 8:38
  • $\begingroup$ I get that idea (more or less). I'm just not sure why we would want to make such a complicated construction. I'm interested in what the advantage is of constructing it this way. $\endgroup$
    – user56834
    Commented Apr 29, 2018 at 8:43
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    $\begingroup$ Complicated is relative term, it is really a basic construction in mathematics (not just linear algebra). The advantage is that $V^*$ exists without choice of base for $V$, while row/column vectors do not. The choice of base is needed only to write an explicit isomorphism between $V$ and $V^*$. $\endgroup$
    – Ennar
    Commented Apr 29, 2018 at 8:56

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This definition is much more general than your suggestion. When $V$ is not a finite-dimensional space the dual space $V^*$ may not be isomorphic to $V$, and not every functional $f: V\to F$ may be represented as some sort of scalar product of two elements of $V$. But in a finite-dimensional case $V^*$ is isomorphic to $V$ and your way of thinking about it is fine.

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