Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a vector space over an arbitrary field $\mathbb{F}$ and denotes its dual by $X^*$. Suppose $k:X\times X^*\to\mathbb{F}$ be a bilinear map.

How can I prove that there exists a linear map $f:X\to X$ such that $k(x,x^*) = x^*(f (x))$ for every $x\in X$ and all $x^*\in X^*$.

share|cite|improve this question
Maybe working with a Hamel basis of $X$? – Davide Giraudo Nov 1 '12 at 11:10
@azimut Maybe useful link: How much bumping is too much? at meta. – Martin Sleziak Mar 7 '13 at 12:13
@MartinSleziak: Thank you. I just wasn't aware that a simple retag bumps the question up to the first page. – azimut Mar 7 '13 at 12:18

The choice of a map $k$ is equivalent to that of a map $k':X\to X^{**}$ satisfying $$ k'(x)(y^*)=k(x,y^*)\qquad\text{for all }x\in X, y^*\in X^*. $$ One has a natural injective map $\iota:X\to X^{**}$ given by $\iota(x)(y^*)=y^*(x)$. The requirement for you map $f$ can now be expressed as $$ k'(x)(y^*)=\iota(f(x))(y^*)\qquad\text{for all }x\in X, y^*\in X^*, $$ which means $\iota(f(x))=k'(x)$ for all $x\in X$, or simply $\iota\circ f=k'$. It can be seen that this is possible if and only the image of $k'$ is contained in that of $\iota$ (in which case one can take $f$ to map any $x$ to the unique $x'\in X$ with $\iota(x')=k'(x)$). Since $k'$ is arbitrary in the question, one cannot guarantee this unless $\iota$ is surjective (hence bijective). Now this is known to be true if and only if $X$ is finite dimensional, so the answer to your question is that only with this additional hypothesis a proof can be given (and the choice of $f$ can be expressed as $f=\iota^{-1}\circ k'$ in this case). In the infinite dimensional case, you can easily give a counterexample once you are given an element of $X^{**}\setminus\iota(X)$, which exists for dimension reasons but may be hard to actually find.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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