- #1
yifli
- 70
- 0
Given that N is an (n-1)-dimensional subspace of an n-dimensional vector space V, show that N is the null space of a linear functional.
My thoughts:
suppose [tex]\alpha_i[/tex]([tex]1\leq i \leq n-1[/tex]) is the basis of N, the linear functional in question has to satisfy f([tex]\alpha_i[/tex])=0.
Am I correct?
Thanks
My thoughts:
suppose [tex]\alpha_i[/tex]([tex]1\leq i \leq n-1[/tex]) is the basis of N, the linear functional in question has to satisfy f([tex]\alpha_i[/tex])=0.
Am I correct?
Thanks