# Thread: The Dual Space and Covectors ... Browdwer, Theorem 12.2 and Corollaries ...

1. I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 12: Multilinear Algebra and am specifically focused on Section 12.1: Vectors and Tensors ...

I need help in fully understanding Corollary 12.4 to Theorem 12.2 ... ...

Theorem 12.2 and its corollaries read as follows:

In the above text from Browder, we read the following:

" ... ... 12.4 Corollary. If $\displaystyle x \in V$, and $\displaystyle x \neq 0$, there exists $\displaystyle \alpha \in V^*$ such that $\displaystyle \alpha (x) \neq 0$ ... ... "

Can someone please demonstrate a formal and rigorous proof for Corollary 12.4 ...?

Help will be appreciated ... ...

Peter  Reply With Quote

2.

3. Hi Peter,

Given a nonzero element $\bf{x}$ of $V$, the set $\{\bf{x}\}$ is linearly independent; this set thus extends to a basis of $V$. By the theorem, the basis has a dual basis in $V^*$ in which one element, call it $\alpha$, satisfies $\alpha(\mathbf{x}) = 1$. In particular, $\alpha(\mathbf{x}) \neq 0$.  Reply With Quote

4. Or put it otherwise, let
$$x=\sum_{j=1}^{n} \xi^j u_j \in V$$
in terms of the basis $\{u_1, \cdots, u_n \}$ of $V$, such that $\alpha(x)=0$ for all $\alpha \in V^*$

Then for each member $\bar{u}^i$ of the basis $\{ \bar{u}^1, \cdots, \bar{u}^n\}$ of $V^*$ we have
$$\bar{u}^i(x)=\xi^i=0$$

Therefore $x=0$

Edit: I changed the positions of some indices.  Reply With Quote

+ Reply to Thread 12.2, 12.4, corollaries, corollary, theorem #### Posting Permissions 