Let $V$ be a vector space over a field $K$, $V^*$ be it's dual (it's linear functionals), $\{\alpha_1,...,\alpha_n\}$ be a basis for $V$ and $\{f_1,...,f_n\}$ be the dual basis. Any subset $S^*$ of the dual basis is linearly independent, i.e.,

$$\sum_{f_i \in S^*} c_i f_i = 0 \iff c_i = 0, \forall i$$

That 0 is the zero function, an element of $V^*$. My question is, let $S$ be a proper subset of the basis of $V$ and $S^*$ a subset of the basis of $V^*$. If I have on $K$:

$$\sum_{f_i \in S^*} c_if_i(a_j) = 0, \forall a_j \in S$$

Does that imply $c_i = 0$ for all $i$? My guess is not, once $S$ is a proper subset of the basis, but I wouldn't be much surprised either if it was true. Either way, I don't know how to show it.

EDIT: As a bonus, if it is false, is there some condition on $S^*$ that would make it true? e.g., $S^*$ is actually the dual of $S$, instead of any subset.

  • $\begingroup$ subset of a linearly independent set is linearly independent $\endgroup$ – Kushal Bhuyan Jun 5 '16 at 2:21
  • $\begingroup$ @KushalBhuyan yes, but my second sum happens not on $V^*$ but on my field $K$, and the $a_j$ doesn't run through all the values on the basis of $V$. Linear independence would follow directly if the sum would happen in $V^*$, or if $a_j$ ran through all the values of the basis, which is not the case. I cannot see a obvious argument in this case, for either answer. $\endgroup$ – Henrique Augusto Souza Jun 5 '16 at 2:24

Consider $S = \{\alpha_1\}, S^*=\{f_2\}$. Then $c_1f_2(\alpha_1)=0$ for any $c_1$. So your hypothesis does not hold generally.

Now take any $S$ and let $S^*$ be its dual. Consider $f=\sum_{f_i\in S}c_i f_i.$ If some $c_j$ is nonzero, then the dual vector $\alpha_j$ of $f_j$ is contained in $S$, and $f(\alpha_j)=c_j\neq 0$.

Hence we see in this case indeed that if $f(\alpha)=0$ for all $\alpha\in S$, then all the $c_i$ are zero.


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