Dual basis explicit form

Question

I'm reading FIS Linear Algebra chapter 2.6, Example 4. Usually when defining a function using direct delta, for example $f(i,j)=\delta_{ij}$ doesn't need anything more.

But why do we need to solve system of equation to find dual basis for a given basis of the vector space? Why does this give a so-called 'explicit form', but not directly from just the dirac-delta definition?

Another question, say by solving the linear system of equations, we obtain $f_1(e_1)=-1$, $f_1(e_2)=-3$, why is it immediate that $f_1(x,y)=-x+3y$?

Answer 1

I don't know what "FIS" is and can't help you with your first question, sorry.

Regarding the second one: $(x,y) = xe_1+ye_2$, so $f(x,y) = f(xe_1+ye_2) = xf(e_1)+yf(e_2)= x(-1)+y(-3) = -x-3y$ (I assume you have a typo in your question). Happy 2018!

Answer 2

It seems you are talking about the following:

$\{e_1,e_2\} $ is a basis of a 2-dimensional space $X$ ; find the corresponding dual basis $\{ f_1,f_2 \}$.

I have never heard of an 'explicit form' for a dual/basis. I presume you mean computing/finding it.

The dirac delta $\delta_{ij}$ is (usually) defined on pairs of integers $i,j$. By itself, it has nothing to do with bases or vector spaces. The dual basis is defined relative to a chosen basis $\{e_1,e_2\}$ of your original vector space $X$ as $$ f_i(e_j) = \delta_{ij}, \quad i,j=1,2$$ which can be understood as a matrix equation, or a system of 4 equations, once one chooses to write things in coordinates.

This equation uniquely determines the linear maps $f_i$. The point of this example (i presume) is for you to figure out what linear maps they are. I don't know how much more 'direct from the dirac-delta definition' you can have than the above equation.

You may know that the dual bases are linear maps. Hence this can be done because linear maps are completely determined by their action on a basis, purely from the linearity (max_zorn's answer).

Perhaps what confuses you is that the 'explicit form' of $f_i$ depends not only on the choice of $e_i$, but also the basis you choose to express them in. That is, $f_i$ is a linear map on $X$. Suppose $e'_1,e'_2$ is a different basis of $X$. Then we can write $f_i$ in this basis too; we will get different numbers out.

Answer 3

It looks like you’ve misinterpreted the expression $f_i(e_j)=\delta_{ij}$ in the definition of dual basis. Take a closer look at the differences between this expression and the function definition $F(i,j)=\delta_{i,j}$ (I’ve replaced $f$ with $F$ for clarity). The arguments to $F$ are a pair of numbers and the expression says that for any inputs, the output of $F$ is identical to that of the Kronecker delta function. Typically, that’s used with natural numbers, so $F: \mathbb N\times\mathbb N \to \{0,1\}$.

On the other hand, if our vector space is $V[K]$ then $f_i:V\to K$. The function takes an element of $V$—a vector—as input and outputs a scalar. It can’t in general be equal to the Kronecker delta function and so the expression $f_i(e_j)=\delta_{ij}$ isn’t in and of itself a function definition. What it does do is tell you the value of $f_i$ for a particular set of inputs, namely the elements of a basis for $V$. The indices in this expression aren’t inputs the to function as they are in the definition of $F$. Instead, they’re part of the names of the objects in the equation. Using indices and the Kronecker delta in this expression allows the $(\dim V)^2$ constraints $$f_1(e_1)=1 \\f_1(e_2)=0\\f_3(e_3)=0 \\ \vdots \\f_2(e_1)=0 \\ f_2(e_2)=1 \\ f_2(e_3)=0 \\ \vdots \\ f_3(e_1)=0 \\ f_3(e_2) = 0 \\ f_3(e_3) = 1 \\ \vdots$$ on the elements of a dual basis to be expressed very compactly. Now, since these are linear operators on $V$, knowing their values on the basis vectors is enough to determine their values on any elements of $V$, so in that sense these are function definitions, but of a very different flavor from $F(i,j)=\delta_{ij}$.