
MHB Master
#1
February 14th, 2019,
23:20
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

February 14th, 2019 23:20
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MHB Master
#2
February 15th, 2019,
00:26
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, $\alpha(\mathbf{x}) \neq 0$.

#3
February 15th, 2019,
04:56
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.
Last edited by steenis; February 16th, 2019 at 00:18.