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suppose we have a vector space $V$ over a field $F$. Then any linear mapping Hom$_F(V,F)$ from $V$ to $V^*$ is also a vector space of $F$. Then the $V^{**:}=\ $Hom$_F(V^*,F)$ is also a vector space over $F$. My confusion comes from studying the canonical homomorphism from $V$ to $V^{**}$, quote from Motivation to understand double dual space :

Let $v \in V$. I am going to build an element $\Phi_v$ of $V^{**}$. An element of $V^{**}$ should be a function that eats functions that eat vectors in $V$ and returns a number.

why is that?

Elements of $V$ are vectors, elements of $V^*$ are all linear mappings, but they are still vectors,

so $V\mapsto V^*$ is going from a vector to a vector, no problem here.

But if that is the case, then why does $V^*\mapsto V^{**}$ now map from a vector to a number? (assuming if we take $F=\mathbb{R}$)

Should $V^*\mapsto V^{**}$ not also just be mapping a vector to another vector?

jeb2
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    This seems to be terminological confusion. "Then any linear mapping [...] is a vector space". No, the set of all these mappings (with appropriate addition and scalar multiplication) is a vector space, not any individual map itself. A "number" in this context is an element of $F$, a "vector" is an element from any vector space over $F$. Since $F$ is a vector space over itself, numbers are also vectors and these are not mutually exclusive concepts. – Thorgott Feb 10 '22 at 23:16
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    Also, I'm not sure what $V\mapsto V^{\ast}$ is supposed to mean. If you're thinking of an operation on $V$, you're just turning one vector space into another. But there is no natural linear map $V\rightarrow V^{\ast}$. Same for $V^{\ast}\mapsto V^{\ast\ast}$. And, in either case, neither is $V^{\ast\ast}$ a number, nor are elements of $V^{\ast\ast}$. Instead, elements of $V^{\ast\ast}$ take elements of $V^{\ast}$ and turn them into numbers. – Thorgott Feb 10 '22 at 23:17
  • @jeb2 You might find this explanation of double dual spaces to be helpful. – Ben Grossmann Feb 11 '22 at 15:01

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