We can identify spaces $V$ and $V^{**}$ by canonical isomorphism: $$A:V\to V^{**},$$ $$Av(f)=f(v),$$ for any $f\in V^*$.

But why we cannot identify $V$ and $V^*$ by $e^{*}_{i}(e_j)=\delta_{ij}$ (I understand that after change the basis of $V$ operator $B: V\to V^*$ will be changed)? What means that spaces $V$ and $V^{**}$ are identical? How we can use it?

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    $\begingroup$ $V$ and $V^{**}$ are not "identical", but in the finite dimensional case they are canonically isomorphic. The isomorphism $A$ does not depend on the choice of basis for $V$, so it is "coordinate-free". Whereas the isomorphism between $V$ and $V^*$ is non-canonical, because it depends on the choice of basis. So $V$ and $V^*$ are also isomorphic, but non-canonically; there is no "natural" or "coordinate-free" isomorphism. $\endgroup$ Jul 13, 2011 at 19:46
  • $\begingroup$ I don't understand how $e_i^{\ast} (e_j) = \delta_{ij}$ defines an isomorphism from $V$ to $V^{\ast}$. $\endgroup$ Jul 13, 2011 at 19:49
  • $\begingroup$ The keyword is canonical. (BTW, you should add "finite dimensional" to your statement.) All $d$ dimensional vector spaces over the same base field $K$ are isomorphic. And the $e_j\mapsto e^*_j$ operation once you fix a basis give an explicit isomorphism. $\endgroup$ Jul 13, 2011 at 19:49
  • $\begingroup$ Isn't your isomorphism $A$ going from $V^**$ to $V$? $\endgroup$
    – Fabian
    Jul 13, 2011 at 19:50
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    $\begingroup$ @Fabian: No: $v\in V$, and $Av\in V^{\*\*}$ so when you apply it to $f\in V^*$, you should get an element of $K$, which you do, namely $f(v)$. He is giving the value of $Av$ at $f\in V^*$, so $Av\in V^{**}$. $\endgroup$ Jul 13, 2011 at 19:52

3 Answers 3


Prof Magidin has answered your question, but I wanted to mention the following exercise: suppose that you give me, for each finite-dimensional vector space $V$ (over $\mathbf{R}$, say), an isomorphism $h_V\colon V \to V^*$. Then you can show that there is some diagram $$ \begin{array}{ccc} V & \xrightarrow{f} & W \\ \downarrow & & \downarrow \\ V^* & \xleftarrow{f^*} & W^* \end{array} $$ (where the vertical arrows are $h_V$ and $h_W$) which does not commute, where $(f^*\lambda)(v) = \lambda(f(v))$ is the usual dual of a linear map. This is not to say that dualizing isn't a functor!


All the other answers are great, but I feel like this simple thing was left unsaid:

The word identical is unwise to use, because it is unclear whether it means isomorphic or naturally isomorphic.

V and V* and V** are all isomorphic of course (if they are finite dimensional), after all they have the same dimension.

What you want to say is this: V and V** are naturally isomorphic but V and V* are not! The notion of natural isomorphism is defined nicely by Dylan. If you look it up you'll find that what is actually naturally isomorphic is not V with V** but rather the ** itself. That is, the functor of double-dual on the category of vector spaces over the reals is naturally isomorphic to the identity functor.


Maybe one should note that this canonical morphism $V \to V^{**}$ is in general not an isomorphism if $V$ is not finite dimensional. Think about the space $V = \oplus_{n\geq 1} K$ for some field $K$. Here already $V^*$ is not isomorphic to $V$. In Banach space theory for example a space is called reflexive precisely if this map is an isomorphism.

I think part of your confusion seems to be that people often say "identical" when they mean canonically isomorphic. If you keep in mind that most of the time things are not "the same" but canonically (or naturally) isomorphic then this will help you not to get confused. In particular the spaces $V$ and $V^{**}$ are not identical.

  • $\begingroup$ It should be remarked that the Banach spaces case usually refers to continuous dual, not algebraic dual. $\endgroup$
    – Asaf Karagila
    Jun 15, 2012 at 7:37

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