1
$\begingroup$

Theorem: Suppose that $V$ is a finite-dimensional vector space wwith the ordered basis $\beta=\{x_1,x_2,\dots ,x_n\}.$ Let $\textbf{f}_i \, (1\leq i \leq n)$ be the $i$th coordinate function with respect to $\beta$ as just defined and let $\beta^* = \{\textbf{f}_1,\textbf{f}_2,\dots ,\textbf{f}_n\}$. Then $\beta^*$ is an ordered basis for $V^*$, and, for any $\textbf{f} \in V^*$, we have

$$\textbf{f}=\sum_{i=1}^n \textbf{f}(x_i)\textbf{f}_i$$

Before I begin the proof I do not understand that very last line. So, I see that $\textbf{f}_i$ are the basis vectors so any vector in the space is a linear combination of the basis vectors but the coefficients $\textbf{f}(x_i)$ throw me off. If these could be explained that would be great.

Proof: Let $f\in V^*$. Since dim$(V^*)=n$, (I assume we are using this for linear independence?) we need only show that $$\textbf{f}=\sum_{i=1}^n \textbf{f}(x_i)\textbf{f}_i$$

from which it follows that $\beta^*$ generates $V^*$, and hence is a basis. Let

$$\textbf{g}=\sum_{i=1}^n \textbf{f}(x_i)\textbf{f}_i$$ (^this feels like assuming the conclusion?)

For $1 \leq j \leq n$, we have

$$\textbf{g}(x_j)=\left(\sum_{i=1}^n \textbf{f}(x_i)\textbf{f}_i\right)(x_j)=\sum_{i=1}^n \textbf{f}(x_i)\textbf{f}_i(x_j)=\sum_{i=1}^n \textbf{f}(x_i)\delta_{ij}=\textbf{f}(x_j)$$

Therefore $\textbf{f}=\textbf{g}$.

I believe the above line but as I noted it feels like we have assumed the conclusion. Where is my thinking going wrong and what is the intuition here?

See related question here

Dual basis existence and uniqueness.

$\endgroup$
1
$\begingroup$

It's right of you to be concerned. Let ${\bf x} \in V$ with ${\bf x} = (\alpha_1, ..., \alpha_n)$. Write ${\bf x} = \sum\limits_{k=1}^n \alpha_i {\bf e}_i$, where ${\bf e}_i$ are the standard coordinates. Then for an element ${\bf f} \in V^\ast$, we have

$$ {\bf f}({\bf x}) = \sum\limits_{k=1}^n \alpha_i\, {\bf f}({\bf e}_i)= \sum\limits_{k=1}^n {\bf f}_i({\bf x}) {\bf f}({\bf e}_i) $$

So, it seems they have written ${\bf f}(x_i)$ for ${\bf f}({\bf e}_i)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.