1

Let $V$ be a finite dimensional vector space over a field $F$.

Let $v\in V$ with $v$ not equal to $0$. Show that there is $\varphi \in V^*$ such that $\varphi(v)$ is not equal to $0$.

I know that $V^*$ is the vector space consisting of all linear functionals on $V$ with the operations of addition and scalar multiplication. Not sure how to start this proof however. Any help would be appreciated. Thanks.

Community
  • 1
kkkk
  • 491
  • Here's a hint. Recall that you can define a linear transformation by its action on a basis. All linear functionals are linear transformations form a vectorspace to its underlying field. Also, I think that you meant to write that v is nonzero. Also, you need V to be a nontrivial vector space, of course – RougeSegwayUser Apr 07 '13 at 21:01
  • I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here. – A.P. Apr 07 '13 at 21:02
  • Hint: Coordinate functions for a basis are examples of linear functions. Can you arrange to bring a suitable one into the picture? – Marc van Leeuwen Apr 07 '13 at 21:04

1 Answers1

4

Hint: Extend $v$ to a basis of $V$, say $\{v=v_1, v_2,v_3,\dots v_n\}$. A function $T:V\rightarrow F$is linear iff $T\bigg(\sum_{i=1}^n\lambda_iv_i\bigg) =\sum_{i=1}^n\lambda_iT(v_i)$ for any $\lambda_i \in F$. Note that $T$ is uniquely determined by its action on a basis, so defining the values of $T(v_i)$ defines $T$ uniquely, and when defining $T$ on a basis, the values can be anything you want.

Tom Oldfield
  • 13,034
  • 1
  • 39
  • 77
  • So if we define v1=v. and let B={x,...,xn} be the dual basis of V. Then x1(v1)=1 not 0. Let x=x1. Therefore, x(v) does not equal 0. – kkkk Apr 09 '13 at 06:54