Let $V$ be a vector space of dimension $n$ over the field $K$. Let $V^{**}$ be the dual space of $V^{*}$. Show that each elelment $v\in V$ gives rise to an element $\lambda_v$ in $V^{**}$ and that the map $v\to\lambda_v$ gives an isomorphism of $V$ with $V^{**}$. Book Linear Algebra, Serge Lang

Theorem: The map $v\to L_v$ of $V$ into $V^{*}$ is an isomorphism.I define $L_v(w)=\langle v,w \rangle$, as the functionals of $V^{*}$

Proof:We know that $\dim V=\dim V^{*}$ and by non-degeneracy of the inner product the kernel of the map is 0.

I tried to see the map from $V\to V^{**}$ as $v\to L_{L_v}=\lambda_v$. By the property of non-degenerancy we have the kernel of the map $v\to L_{L_v}=\lambda_v$ must be zero. $L_v(w)=0$ so it is $L_{L_v}(w)=0$, for $w\in V$.If we consider $v_1,v_2...v_n$ as a basis that generate $V$ we have L_{v_1}...L_{v_n} as the basis of $V^{*}$. We can apply the linearity of the linear functionals and we get $L_{a_1L_{v_1}...a_nL_{v_n}}=a_1L_{L_{v_1}}+...a_nL_{L_{v_n}}=a_1\lambda_{v_1}+...a_2\lambda{v_n}$. If we consider $a_1v_1+...a_nv_n=0$ when $a_1=...a_n=0$.

$L_{a_1L_{v_1}...a_nL_{v_n}}=a_1L_{L_{v_1}}+...a_nL_{L_{v_n}}=a_1\lambda_{v_1}+...a_2\lambda{v_n}=0$ which means $\dim V=\dim V^{**}$. Then However I am not seeing how I am going end this proof. I am self-studying.

For those who do not understand my terminology and the way I thought, here it is the proof in which I based my own for this exercise.

Theorem: Let V be a finite dimensional vector space over $K$, with a non-degenerate scalar product. Given a functional $L:V\to K$ there exists a unique element $v\in V$ such that: $L(w)=\langle v,w\rangle$

for all $w\in V$.

Proof. Consider the set of all functionals on $V$ which are of type $L_v$, for some $v\in V$. This set is a subspace of $V*$, because of the zero functional is of this type, and we have the formulas


Furthemore, if $\{v_1,...,v_n\}$ is a basis of $V$, then $L_{v_1},...L_{v_n}$ are linearly independent. Proof: If $x_1,...,x_n\in K$ are such that:

$x_1L_{v_1}+...+x_nL_{v_n}=0\\L_{x_1v_1}+...+L_{x_n v_n}=0$

and hence

$L_{x_1v_1+...+x_n v_n}=0$

However, if $v\in V$, and $L_v=0$, then $v=0$ by the definition of non-degeneracy. Hence:

$x_1v_1+...+x_n v_n=0$,

and therefore $x_1=...=x_n=0$, thereby proving our assertion. We conclude that the space of functionals of type $L_v\:(v\in V)$ is a subspace of $V*$, of the same dimension as $V*$, whence equal to $V*$. This proves the theorem.$\blacksquare$ Book: "Linear Algebra" by Serge Lang


Could someone prove this?

Thanks in advance!


I don't know how you define $L_v$: it appears that you're assuming the existence of an inner product, but this is not the case. For instance, nondegenerate bilinear forms may not exist over some field $K$, whereas the isomorphism in the exercise can be defined over any field and any finite dimensional vector space over it.

The map Lang has in mind is $$ \lambda\colon V\to V^{**},\qquad v\mapsto\lambda_v $$ where, for $\varphi\in V^{*}$, $$ \lambda_v(\varphi)=\varphi(v) $$ Note that $\lambda_v$ should belong to $V^{**}=(V^*)^*$, so it should be a linear map $V^*\to K$ and this $\lambda_v$ satisfies the requirements (check it).

It's rather easy to show $\lambda$ is linear. It is injective because for every $v\in V$, if $v\ne 0$ there exists $\varphi\in V^*$ with $\varphi(v)\ne0$.

Finally, if $V$ is finite dimensional, then $$ \dim V=\dim V^*=\dim (V^*)^*=\dim V^{**} $$ and the rank-nullity theorem allows you to finish.

  • $\begingroup$ $L_v$ is a linear functional. I got it from Lang´s. $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:07
  • $\begingroup$ @PedroGomes Useless, if you don't say what it is. But you need not go through $V^*$. $\endgroup$ – egreg Jul 8 '17 at 16:09
  • $\begingroup$ It's at times like this that I wish lambda notation for defining functions were standard mathematical notation. Then you could define $\epsilon : V \to V^{**}$ as $\epsilon := \lambda (v \in V) . (\lambda (\phi \in V^*) . \phi(v))$ (with the understanding that the implicit conversion from $V^* \to k$ to $V^{**}$ requires proving the term is linear for each $v \in V$). $\endgroup$ – Daniel Schepler Jul 8 '17 at 16:10
  • $\begingroup$ $L_v(w)=\langle v,w \rangle$, since it is a linear functional. So I am using it twice to get into $V^{**}$ $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:11
  • 2
    $\begingroup$ It might be an interesting exercise to show if you choose a basis of $V$, and then use the dual basis of $V^*$, then the map $v \mapsto L_{L_v}$ is exactly equal to egreg's map $\lambda$. $\endgroup$ – Daniel Schepler Jul 8 '17 at 16:13


This is independent of the choice of any basis. Consider, for each $v\in V$ the map: \begin{align} V^*&\longrightarrow K&&(\text{$K$ is the base field})\\ \varphi_v\colon f&\longmapsto f(v) \end{align} Show this defines a linear form on $V^*$, i.e. an element of $V^{**}$, and prove the map $\;\varphi\colon v\mapsto \varphi_v$ is a linear injective map.

Some details:

  • ‘$\varphi_v$ is a linear form on $V^{*}$’ means that, for all $f,g\in V^{*}$, one has $$\varphi_v(f+g)=\varphi_v(f)+\varphi_v(g)\quad\text{and}\quad \varphi_v(\lambda f)=\lambda \varphi_v(f)$$ Indeed, $\;\varphi_v(f+g)\stackrel{\text{def}}{=}(f+g)(v)=f(v)+g(v)\stackrel{\text{def}}{=} \varphi_v(f)+\varphi_v(g)$. Similarly for the other relation.
  • Saying that $\varphi$ is linear (w.r.t. $v$) means you have to check $$\varphi_{v+w}=\varphi_v+\varphi_w,\quad \varphi_{\lambda v}=\lambda\varphi_v \qquad\text{for all } v,w\in V,\;\lambda\in K.$$ I'll show the second relation: it means that for all $f\in V^*$, one has $\varphi_{\lambda v}(f)=\lambda\varphi_v(f) $. This is obvious: $$\varphi_{\lambda v}(f)\stackrel{\text{def}}{=} f(\lambda v)= \lambda f(v)\stackrel{\text{def}}{=} \lambda\varphi_v(f). $$ Similarly for the first relation.
  • To prove injectivity, note that $$\ker\varphi=\{v\in V\mid \varphi_v(f)=0\quad\forall f\in V^*\}.$$ $v\in\ker \varphi$ thus means $f(v)=0$ for all $f\in V^*$. This is impossible if $v\ne 0 $ since we can extend $\{v\}$ to a basis of $V$ and if we denote $p_1$ the first coordinate map for this basis, we have by definition $p_1(v)=1$. This proves $\ker\varphi=\{0\}$, so $\varphi$ is injective.
  • Finally, injectivity implies bijectivity since $V$ and $V^{**}$ have the same dimension.
  • $\begingroup$ The proofs I have been through make use of a basis. They pick up one. Why are you saying we cannot go that way? $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:29
  • $\begingroup$ I didn't say you can't go that way. Simply it's more elegant when you don't have to use one. This is a canonical isomorphism i.e. it does not depend on the choice of any basis. $\endgroup$ – Bernard Jul 8 '17 at 16:31
  • $\begingroup$ Thanks for the reply! So could you complete your proof that way so that I can learn it? $\endgroup$ – Pedro Gomes Jul 8 '17 at 16:33
  • $\begingroup$ I can give some more details, but I'd prefer you try to check them by yourself first. $\endgroup$ – Bernard Jul 8 '17 at 16:37
  • $\begingroup$ For example, a basis-independent formulation of $\lambda$ makes it easier to prove: if $T \in L(V, W)$ is a linear transformation, and $\lambda_V : V \to V^{**}$ and $\lambda_W : W \to W^{**}$ are the corresponding maps, then $\lambda_W \circ T = T^{**} \circ \lambda_V$ where $T^{**} : V^{**} \to W^{**}$ is the dual to $T^* : W^* \to V^*$. $\endgroup$ – Daniel Schepler Jul 8 '17 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.