2
$\begingroup$

So we have our definition of the dual basis:

We call the ordered basis $\beta^*=\{\textbf{f}_1,\dots ,\textbf{f}_n\}$ of $V^*$ that satisfies $\textbf{f}_i(x_j)=\delta_{ij}$ the dual basis of $\beta$.

Take this example:

The space of square $n \times n$ matrices. We have an example of a linear functional $\textbf{f}:V \to \mathbb{R}$ that maps a matrix $A$ to $tr(A)$. (an example functional just to demonstrate my understanding). What would be the dual basis to the standard basis for this space?

Another example $\mathbb{R}^2$:

Let $\beta =\{(2,1),(1,3)\}$. We wish to find the dual basis $\beta^*=\{\textbf{f}_1,\textbf{f}_2\}$. To explicitly determine a formula for $\textbf{f}_1$ we consider

$1=\textbf{f}_1(2,1)=\textbf{f}_1(2e_1+e_2)=2\textbf{f}_1(e_1)+\textbf{f}_1(e_2)$ $0=\textbf{f}_1(1,3)=\textbf{f}_1(e_1+3e_2)=\textbf{f}_1(e_1)+3\textbf{f}_1(e_2)$

Why do we use the standard basis here instead of the given basis explicitly?

After all this is done, does the dual basis always induce a natural inner product? If so, what does that really mean intuitively and what does this do for us?

$\endgroup$
1
$\begingroup$

If you have a basis $\mathcal B$ of a vector space $V$, the dual basis $\mathcal B^*$ of $V^*$ simply consists in the (parallel) projections on the vectors of $\mathcal B$.

Thus, in the example you mention, the standard basis is the set of matrices $E_{ij}=(e_{kl})$, where $e_{kl}=0$ if $(k,l)\ne (i,j)$, $e_{ij}=1$. Its dual basis is made up of the maps $$A=(a_{kl})\longmapsto a_{ij}.$$

Second question: the standard basis is used just because all cordinates are given in the standard basis… Note that since $f_1=2e_1+e_2$, we have $f_1^*=2e_1^*+e_2^*$.

Last question: no, the dual basis does not induce an inner product, because this notion is valid for vector spaces over abstract fields, while inner products are deined for real (or complex) vector spaces. However, for real vector spaces the notions are linked through a natural pairing: \begin{align*} V\times V^*&\to \mathbf R,\\ (x,u)&\mapsto u(x). \end{align*}

$\endgroup$
6
  • $\begingroup$ The basis was not given as the standard basis? $\endgroup$ – RhythmInk Nov 2 '15 at 21:23
  • $\begingroup$ Do you mean in your second example? The standard (or rather canonical) basis is $\;\{(1,0),(0,1)\}$. $\endgroup$ – Bernard Nov 2 '15 at 21:26
  • $\begingroup$ That is correct. $\endgroup$ – RhythmInk Nov 2 '15 at 21:27
  • $\begingroup$ Well, $\;\{(2,1), (1,3)\}\neq \{(1,0),(0,1)\}$. I don't really see what is your problem. $\endgroup$ – Bernard Nov 2 '15 at 21:30
  • $\begingroup$ What is throwing me as why we bothered to switch back to the standard basis for our calculation? Is there not a way to do it with any arbitrary basis without making reference to the canonical basis? $\endgroup$ – RhythmInk Nov 3 '15 at 4:27

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.