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 $V^*$.



Prove that $\ker \psi =0.$

  • 1
    $\begingroup$ 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 $v \in V$ $\endgroup$ Apr 1 '20 at 15:35
  • $\begingroup$ No. You have to prove that if $\varphi(x)=\varphi(y)$ then $x=y.$ $\endgroup$
    – Leox
    Apr 1 '20 at 15:40
  • $\begingroup$ 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? $\endgroup$ Apr 1 '20 at 15:43
  • $\begingroup$ Yes, just show that the equation $\varphi(x)=0$ has only trivial solution $\endgroup$
    – Leox
    Apr 1 '20 at 15:46
  • $\begingroup$ Thanks for the clarification. $\endgroup$ Apr 1 '20 at 15:48

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