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Assume that V is finite dimensional vector space over $\mathbb{K}$. Let $<-,->$ denote an inner on V.

Let $D : V \rightarrow V^{*}$ defined as $$v \mapsto D(v)= \ <v,-> \ : V \rightarrow \mathbb{K}$$

It is clear that D is linear. Assume that $D(v) = 0 \implies <v,->$ is the zero map.

In particular $<v,v> = 0 \implies \|v\| = 0 \implies v = 0$

I am having troubles proving that D is surjective.

joe blacksmith
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    For $v\neq 0$, $\langle v,v\rangle\neq 0$, so $\langle v,\alpha v\rangle$ takes all possible values as $\alpha$ ranges over your field. – Arturo Magidin Nov 21 '20 at 01:12
  • This makes sense. Thank you. – joe blacksmith Nov 21 '20 at 01:14
  • If someone adds the details I will accept the answer. – joe blacksmith Nov 21 '20 at 01:16
  • Though I guess what I did was show $D(v)$ is surjective for $v\neq 0$, rather than showing that we get all functionals. That we get all functionals depends on showing that $V^$ has the same dimension as $V$; take an orthonormal basis for $V$, and show that the image in $V^$ is linearly independent to conclude $D$ is onto. Note the result is false if $V$ is not finite dimensional, so you need to use finite dimensionality somewhere. – Arturo Magidin Nov 21 '20 at 01:29
  • P.S. User \langle and \rangle, not < and >. – Arturo Magidin Nov 21 '20 at 01:35

2 Answers2

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For a finite dimensional inner product space $V$, we know that $V^*$ has dimension $\dim(V)\dim(K) =\dim(V)$. To show that $D$ is onto, it suffices to exhibit a linearly independent subset of $\mathrm{Im}(D)$ with $\dim(V)$ elements.

To that end, let $\mathbf{v}_1,\ldots,\mathbf{v}_n$ be an orthonormal basis for $V$. Then note that $D(\mathbf{v}_i)(\mathbf{v}_j) = \delta_{ij}$ (Kronecker’s delta). If $\alpha_1D(\mathbf{v}_1)+\cdots+\alpha_nD(\mathbf{v}_n) = \mathbf{0}$, then evaluating at $\mathbf{v}_i$ we conclude that $\alpha_i=0$. Thus, these images are linearly independent, and hence form a basis for $V^*$. This shows $D$ is surjective.

We need to use finite dimensionality, since the algebraic dual of an infinite dimensional vector space is not isomorphic to the space (canonically or otherwise). See here for a proof. The finite dimensionality occurs twice in the above proof: the existence of an orthonormal basis, and the dimension of $V^*$.

Arturo Magidin
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Let $L \in V^{*}$. Choose orthonormal basis {e_1,\ldots,e_n} of V.

$$L(v) = L(x_1e_1 + \ldots + x_ne_n) = x_1L(e_1) + \ldots x_nL(e_n)$$

Let $v^{\prime} = e_1 L(e_1) + \ldots e_n L(e_n)$. It follows that

$$D(v^{\prime})(v) = D(e_1 L(e_1) + \ldots e_n L(e_n))(v) = \langle e_1 L(e_1) + \ldots e_n L(e_n), x_1e_1 + \ldots x_ne_n> = x_1L(e_1) + \ldots x_nL(e_n) \rangle = T(v)$$

joe blacksmith
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