# Linear Functionals Inner Product

1. Dec 10, 2006

### wurth_skidder_23

Assume that $$m<n$$ and $$l_1,l_2,...,l_m$$ are linear functionals on an n-dimensional vector space
$$X$$.

Prove there exists a nonzero vector $$x$$ $$\epsilon$$ $$X$$ such that $$< x,l_j >=0$$ for $$1 \leq j \leq m$$. What does this say about the solution of systems of linear equations?

This implies
$$l_j(x)$$ $$\epsilon$$ $$X^\bot$$ for $$1 \leq j \leq m$$ or $$l_j(x)=0$$ for $$1 \leq j \leq m$$. Since it is stated in the problem that $$l_1,l_2,...,l_m$$ are linear functionals on the vector space X, $$l_j(x)=0$$. Does this reasoning even help me find the proof? I am stuck.

If you have trouble reading this, it is also at http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/olatex_33882.pdf

Last edited: Dec 10, 2006
2. Dec 10, 2006

### StatusX

I don't know what $X^\perp$ means, nor why you are taking the inner product rather than just applying the functional to the vector. You'll obviously need to involve n and use n>m, and the easiest way to use this is appeal to the existence of a basis with n elements.

3. Dec 10, 2006

### wurth_skidder_23

so basically I'm trying to prove that for some nonzero $$x$$, $$l_j(x)=0$$?

4. Dec 10, 2006

### StatusX

Yea, for some x, for all j.