Inner product gives vector space isomorphism with dual space

Let $$V$$ be a finite dimensional vector space over the field $$\mathbb{K}$$, then $$V$$ is isomorphic to its dual $$V^*$$.

To see this let $$B=\{e_1,...e_n\}$$ is a basis for $$V$$. Then any vector $$v \in V$$ can be uniquely written as $$v=\alpha_1 e_1+...+\alpha_n e_n$$.

Now define $$e^i:V \to \mathbb{K}, e^i(v)=\alpha_i$$, then $$B^*=\{e^1,...e^n\}$$ is a basis for $$V^*$$

If we now define a function $$f:B \to V^*$$ s.t. $$f(e_i)=e^i$$ for all $$i$$, then there exists a unique linear map $$T:V \to V^*$$ with $$T(e_i)=f(e_i)=e^i$$.

To show that $$T$$ is an isomporhism we need to show that $$T$$ is bijective.

To see that $$T$$ is surjective let $$v^* \in V^*$$. Then $$v^*$$ can be uniquely written as $$v^*=\gamma_1 e^1+...+\gamma_n e^n$$ and $$e^i=T(e_i)$$ for all $$i$$, so $$v^*=\gamma_1 T(e_1)+...+\gamma_n T(e_n)=T(\gamma_1 e_1+...+\gamma_n e_n)$$. Since $$\gamma_1 e_1+...+\gamma_n e_n \in V$$, $$v^*=T(v)$$ for some $$v \in V$$ holds for all $$v^* \in V^*$$.

To show that $$T$$ is injective let $$v,w \in V$$, $$v=\alpha_1 e_1+...+\alpha_n e_n$$ and $$w=\beta_1 e_1+...+\beta_n e_n$$. Then $$0=T(v_1)-T(v_2)=T(v_1-v_2)=T((\alpha_1-\beta_1)e_1+...+(\alpha_n-\beta_n)e_n)=(\alpha_1-\beta_1)T(e_1)+...+(\alpha_n-\beta_n)T(e_n)=(\alpha_1-\beta_1)e^1+...+(\alpha_n-\beta_n)e^n)$$.

But $$B^*$$ is a basis, so $$\alpha_i-\beta_i=0$$ which implies that $$v=w$$.

Now as mentioned in this other post this isomorphism $$T$$ depends on the choice of basis, but if $$V$$ is an inner product space, then there is a canonical isomorphism defined by the inner product $$g:V \times V \to \mathbb{K}$$.

To see this write $$\psi_v(w)=g(v,w)$$, then $$\psi:V \to V^*$$ maps $$v$$ to $$\psi_v:V \to \mathbb{K}$$ that does not depend on the choice of any basis.

Now my question is how we can show that $$\psi$$ is indeed an isomorphism. Obviously, it is a linear map, but we also need to show that it is bijective. Can someone give me a hint on how to do it? My idea was to somehow use an orthonormal basis and the fact that the inner product of any vector with the i-th element of an orthonormal basis gives the ith coordinate of that vector. Then we should be able to use the fact that $$e^1,...e^n$$ is a basis of $$V^*$$.

Hint

Prove that $$\ker \psi =0.$$

• Thanks, will try that. Just to make sure I understand how to approach this in the given context. We need to prove that only the zero vector $0_V \in V$ is mapped to the zero vector $0_{V^*} \in V^*$ which is the zero map $0_{V^*}: V \to \mathbb{K}, 0_{V^*}(v)=0$ for all $v \in V$ Apr 1 '20 at 15:35
• No. You have to prove that if $\varphi(x)=\varphi(y)$ then $x=y.$
– Leox
Apr 1 '20 at 15:40
• Well, your comment says that I need to prove injectivity I think. And since $V$ and $V^*$ have the same dimensions this is equivalent to n$ullity(\psi)=0$, isn't it? Apr 1 '20 at 15:43
• Yes, just show that the equation $\varphi(x)=0$ has only trivial solution
– Leox
Apr 1 '20 at 15:46
• Thanks for the clarification. Apr 1 '20 at 15:48