Dual Vector Space embedding

Is there an embedding of any vector space $V$ into $V^*$?

As far as I know it is not true. The statement that I know of is that there is natural embedding of $V$ into $V^{**}$

Is there any condition/ special vector spaces for which $V$ can be into $V^*$ naturally?

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In the first line you ask for an embedding, in the last line for a natural embedding. These are different questions (and have different answers). – Hagen von Eitzen Apr 17 '14 at 11:58

If $V$ has an inner product $\langle \cdot,\cdot\rangle$, then for each $v\in V$ we can construct an element $f_v\in V^*$ by $$f_v(x) = \langle x,v\rangle \quad x\in X.$$ This yields a natural embedding of $V$ in $V^*$ for inner product spaces.

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You have such a thing in spaces with an inner product. A inner product $\langle x,y\rangle$ is an inner product on a vector space $V$ over field $K$, i.e. a map $$\langle \cdot,\cdot \rangle \,:\, V\times V \to K$$ that is linear in it's first argument, and either linear or conjugate linear (i.e. $\langle x,\lambda y \rangle = \bar\lambda \langle x,y\rangle$) in it's second argument.

If $\langle \cdot, \cdot \rangle$ is bilinear (respectively conjugate linear in it's second argument), then a linear (respectively conjugate linear) embedding of $V \to V^*$ is given by $$\cdot^* \,:\, V \to V^* \,:\, c \mapsto \left(x \mapsto \langle x,c \rangle \right)$$ If $\langle \cdot, \cdot \rangle$ is bilinear, it's clear that $x^*$ is a linear map $V \to K$ for every $x$, and that $\cdot^*$ is linear as a map $V \to V^*$. For it to be injective, it must hold that $x^*$ is the zero map only if $x$ is the zero vector. This is e.g. the case if $\langle \cdot,\cdot \rangle$ is positive definite, because then for every non-zero $x$, $x^*(x) = \langle x,y \rangle > 0$. For generally, it suffices for the injectivity of $\cdot^*$ that for every $x$ there is an $y$ with $\langle x,y \rangle \neq 0$. If $\langle \cdot, \cdot \rangle$ is no bilinear, but rather conjugate linear in it's second argument, then the construction above yields a conjugate linear map $V \to V^*$.

Conversely, if you have an embedding $\cdot^* \,:V \to V^*$, you can define an inner product by $$\langle \cdot,\cdot \rangle \,:\, V\times V \to K \,:\, \langle x,y \rangle = y^*(x) \text{.}$$ Note, however, that in the general case this inner product might not be positive definite, and not (conjugate) symmetric.

If we drop all pretense of naturality, talk pure vector spaces only, i.e. $V^*$ means the algebraic dual of $V$, and assume the axiom of choice (AC), we can do the following:

Let $V$ be a vector space over $K$, and $B$ be a basis of $V$ - the existence of which is guaranteed by AC. Then, by the very definition of vector space basis, there's a coordinatization map $$\mathfrak{C}_B \,:\, V \to K^B \,:\, x \mapsto (x_b)_{b \in B}$$ such that for all $x \in V$ only finitely many $x_b \neq 0$ and $$x = \sum_{b \in B} x_b b \text{.}$$ Note that $\mathfrak{C}_B$ is linear, if we interpret sums and products on $K^B$ pointwise, i.e. we have that $\left(\mathfrak{C}_B(x+y)\right)_b = \left(\mathfrak{C}_B(x)\right)_b + \left(\mathfrak{C}_B(y)\right)_b$ and also $\left(\mathfrak{C}_B(\lambda x)\right)_b = \lambda\left(\mathfrak{C}_B(x)\right)_b$ for all $b \in B$.

We can now define a inner product on $V$ just as we do in the finite-dimensional case, i.e. set $$\langle \cdot,\cdot \rangle \,:\, V\times V \to K \,:\, \langle x,y \rangle = \sum_{b \in B} x_b \overline{y_b} \quad\text{where (x_b)_{b \in B} = \mathfrak{C}_B(x) and (y_b)_{b \in B} = \mathfrak{C}_B(y).}$$ (If the underlying field is not $\mathbb{C}$, replace $\overline{y_b}$ with just $y_b$). That sum is well-defined because if only finitely many $x_b\neq 0$ and $y_b \neq 0$, then also only finitely many summands $x_by_b \neq 0$. The linearity of $\langle \cdot,\cdot \rangle$ in both argument follows from the linearity of $\mathfrak{C}_B$. If the underlying field $K$ is $\mathbb{R}$ or $\mathbb{C}$ this inner product is positive definite, because then for $x \neq 0$ you have $\sum_{b\in B} x_b\overline{x_b} = \sum_{b \in B} |x_b|^2 > 0$. If, on the other hand, $K = \textrm{GF}(p^n)$, then $\langle x,x \rangle = 0$ if $x = b_1 + \ldots b_n$ for $n$ distinct $b_1,\ldots,b_n \in B$.

And now, just as above, we have an embedding $V \to V^*$, defined by $$\cdot^* \,:\, V \to V^* \,:\, c \mapsto \left(x \mapsto \langle x,c \rangle \right)$$ If $x \in V$, $x \neq 0$ and $(x_b)_{b \in B} = \mathfrak{C}_B(x)$ is the coordinatization of $x$, then $x_{b} \neq 0$ for some $b$. Then $x^*(b) = x_b \neq 0$, i.e for every non-zero vector $x$, $x^*$ is not the zero map, which proves that $\cdot^*$ is injective (even if the inner product was not positive definite).

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if it is not inner product space, is it ever (not necessarily naturally) possible to find a basis for $V$, such that there exists an embedding? More precisely, is it possible to define a basis of $Hom(U,V)$ such that there exists embedding of $Hom(U,V)$ into $Hom(U,V)^*$. I am repeating(perhaps) this, as I was assigned this as homework and I can't see how. – user143774 Apr 17 '14 at 11:01
You'll have to define exactly what you require of an embedding. If you assume the axiom of choice, both $V$ and $V^*$ have a basis, say $B$ and $B^*$. If these have the same cardinality, there's always a linear bijection $V \to V^*$, and if $|B| \leq |B^*|$ there's always a linear injection $V \to V^*$. But these are, in general, very far from "natural", I'd say. – fgp Apr 17 '14 at 11:15
I believe naturally, embedding (in context of vector spaces) would be injective linear map. – user143774 Apr 17 '14 at 11:29
@user143774 I've extended my answer to show that such an embedding always exists if you assume the axiom of choice. Note, however, that this construction is hightly non-constructive because you usually don't know much about a basis other than that it exists. In particular, that embedding won't usually be compatible with any additional structure you have on $V$ like a norm or a topology. – fgp Apr 17 '14 at 11:55
When you say "It's clear that $⋅^∗$ is linear", I would say that this is only clear if you take the "inner product" to be bilinear (but then in the complex setting is is not an inner product); if instead it is conjugate-linear in its second argument, then the map $⋅^∗$ would seem to be conjugate-linear too. – Marc van Leeuwen Apr 17 '14 at 13:41