25 - Inner Product-Dual Isomorphism

Inner Product-Dual Isomorphism

Let $V$ be an inner product space over $\mathbb{K}=\mathbb{R},\mathbb{C}$ then
For all $v\in V$

$$\begin{array}{rl}⟨v,\cdot ⟩:V& \to \mathbb{K}\\ \text{ in other words }w& \mapsto ⟨v,w⟩\end{array}$$

is a linear functional as $⟨,⟩$ is linear in the second co-ordinate

Map $v\mapsto ⟨v,\cdot ⟩$ is a natural injective $\mathbb{R}-$linear map

$$\varphi :V\to V^{\prime }$$

which is an isomorphism when $V$ is finite dimensional

Note that every complex vector space $V$ is a real vector space and if it is finite dimensional then

$$2{\dim }_{\mathbb{C}}(V)={\dim }_{\mathbb{R}}(V)$$

Proof

Note $\varphi :v\mapsto ⟨v,\cdot ⟩$ so need to show $\varphi (v+\lambda w)+\varphi (v)+\lambda \varphi (w)$ for all $v,w\in V$, $\lambda \in \mathbb{R}$
So

$$⟨v+\lambda w,\cdot ⟩=⟨v,\cdot ⟩+\lambda ⟨w,\cdot ⟩$$

which is true by linearity of an inner product space
Hence $\varphi$ is $\mathbb{R}-$linear (conjugate linear for $\lambda \in \mathbb{C}$ )

As $⟨\cdot ,\cdot ⟩$ is non-degenerate then

$$⟨v,\cdot ⟩=\overline{⟨\cdot ,v⟩}\text{ is not the zero functional unless }v=0$$

Hence $\varphi$ is injective

If $V$ is finite-dimensional then ${\dim }_{\mathbb{R}}V={\dim }_{\mathbb{R}}{V}^{\prime }$ hence $\mathrm{Im}\varphi =V^{\prime }$
So $\varphi$ is surjective and hence

$$\varphi \text{ is a }\mathbb{R}\text{-linear isomorphism}$$