Let $V$ be a infinite dimensional vector space over a field $K$, and $\{v_i\}_{i\in I}$ be a basis of $V$. For each $i\in I$, let $f_i: V\to K$ be defined by $f_i(v_j)=\delta_{ij}$. Prove that $\{f_i\}_{i\in I}$ is linearly independent but does not span the dual space $V^*$.

I can prove that $\{f_i\}_{i\in I}$ is linearly independent. But I cannot find a counterexample in $V^*$ that is not spanned by $\{f_i\}_{i\in I}$. Does anyone has any idea on how to create this counterexample? This is the difference between finite and infinite dimensional vector space, right?

Thank you very much!

  • 4
    $\begingroup$ Remember that linear combinations only have finitely many non-zero coefficients, even in infinite-dimensional vector spaces. $\endgroup$ – celtschk Mar 21 '15 at 21:30
  • $\begingroup$ Aha~~Thank you for your remind! Let me think about this. $\endgroup$ – breezeintopl Mar 21 '15 at 21:31

Let $V$ the real vector space generated by an infinite set $A$. By definition, $$V:=\{f:A\rightarrow\mathbb{R}\,\mbox{ s.t. } f(a)\neq 0\mbox{ for finitely many }a\in A\}$$ A basis of $V$ is $\{e_a\}_{a\in A}$, where $e_{a_i}(a_j)=\delta_{i,j}$. Let's denote by $e_a'$ the corresponding elements of $V^*$, i.e. $e_{a_i}'(e_{a_j})=\delta_{i,j}$. Now consider $\psi\in V^*$, defined by $$\psi(f):=\sum_{a\in A}f(a)$$ Observe that $\psi$ is well defined cause the sum involves only a finite number of terms. Suppose now that $\psi$ can be written as a (finite) linear combination of $e_a'$. Hence $$\psi=\sum_{a\in B}\lambda_ae_a'$$ where $B$ is a finite subset of $A$, and $\lambda_a$ are scalars. Taking $\bar{a}\not\in B$ (it exists cause $A$ is infinite while $B$ is finite), and applying the last identity to $e_\bar{a}$, we get $1=0$, contradiction. In general, $V^*\equiv \mathbb{R}^A$ (the set of all real-valued functions on $A$), while the space spanned by $\{e_a'\}_{a\in A}$ is isomorphic to $V$ (the set of real-valued functions on $A$ which are $0$ for all but finitely many $a\in A$).

  • $\begingroup$ So "span" can only be defined for the sum of finite number of terms(even for infinite dim space)? $\endgroup$ – breezeintopl Mar 23 '15 at 23:06
  • $\begingroup$ Yes, that's the point! $\endgroup$ – Capublanca Mar 23 '15 at 23:41

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.