How to compute a dual basis

Question

I have been searching everywhere for good resources (including other questions here) for understanding how to compute a dual basis given a basis in a vector space. Every single answer I found is confusing/doesn't explain how to actually work out a problem in full. Could somebody please help me understand the answer to the problem listed below. Where do these systems of equations that seem to spring out of nowhere come from? I understand it has something to do with the delta function but I am lost as to what the actual computation of going from the basis to the system of equations is. Please explain it in simple but thorough terms.

Answer 1

Let's just focus on the first system, and use $e_j'$ to denote the dual basis elements (not standard, but it's used in the problem). We want to find the coordinates of $e_1'=(x_1\ y_1 \ z_2)$ so that $e_1'(e_1)=1,\ e_1'(e_2)=0,$ and $e_1'(e_1)=0$. You can either see this through using the standard basis and its dual basis or doing $e_1'e_j$ as row-column vector multiplication.

This gives the system of equations $$\begin{align*}x_1+y_1+2z_1&=1\\x_1+z_1&=0\\2x_1+y_1&=0. \end{align*}$$ Solving this system gives us $e_1'.$ Now, do the same for the other ones.

EDIT: I'll expand on computing, say, $e_1'(e_1)$. As I said, one way is to do the multiplication of these vectors ($e_1'$ is row and $e_1$ is a column). That's pretty straightforward and represents one way to get the first equation in the system.

Alternatively, we can write $e_1=f_1+f_2+2f_3$ and $e_1'=x_1f_1'+y_1f_2'+z_1f_3'$ where $(f_j)$ and $(f_j)'$ denote the standard basis and its dual basis (respectively). Since $f_i'(f_j)=\delta_{ij}$, we know that $x_1f_1'(f_1)=x_1,\ x_1f_1'(f_2)=0,\ x_1f_1'(f_3)=0$, and then similarly for the other parts of $e_1',$ which will give us the first equation in the system.

Both of these ways are doing the same thing, but one may be more understandable for you right now. Eventually, you want to see these as the same.

Answer 2

The writer seems to use both the notation $e_i'$ and $e^i$ for the dual of $e_i$. Maybe this is the source of your confusion? The three systems of equations come from computing $e^je_i$ or $e_j'e_i$ (which is the same) for $j = 1,2,3$.

Answer 3

The “delta function” that appears to be puzzling you is the Kronecker delta, which is defined by $$\delta_{ij} = \begin{cases}1 & \text{if }i=j, \\ 0 & \text{otherwise}.\end{cases}$$ Another way of putting this is that $\delta_{ij}$ is the $(i,j)$-th element of the identity matrix.

If we write the given basis vectors as rows of a matrix $B$ and the dual basis vectors as columns of $B'$, then the set of defining equations $e_ie^j=\delta_{ij}$ can be written as the single matrix equation $BB'=I$, from which $B'=B^{-1}$. Note that this only works because the bases in which both sets of vectors are expressed are themselves dual: the standard basis is its own dual.

In particular, for the basis in your question, $$B=\begin{bmatrix}1&1&2\\1&0&1\\2&1&0\end{bmatrix}$$ and $$B^{-1} = \begin{bmatrix}-\frac13&\frac23&\frac13\\\frac23&-\frac43&\frac13\\\frac13&\frac13&-\frac13\end{bmatrix},$$ therefore $$e_1' = \left(-\frac13, \frac23, \frac13\right) \\ e_2' = \left(\frac23,-\frac43,\frac13\right) \\ e_3' = \left(\frac13,\frac13,-\frac13\right).$$

More generally, vectors aren’t always simply tuples of scalars, and their duals are more generally linear functionals—functions that take a vector and produce a scalar. You might not have a convenient pair of dual bases already so that you can use coordinate vectors to do this calculation, in which case you’ll need to fall back on the defining equation of a dual basis develop a system of equations for the dual basis. However, when you can express vectors and their duals as coordinates as above, then computing a dual basis amounts to finding the inverse transpose of a matrix.

Answer 4

A last possibility would be using the cross product to build the dual basis:. Indeed the vectors defined as $$\left(\frac{v_i\times v_j}{(v_i\times v_j)\cdot v_k}\right)^t\quad i\neq j\neq k$$ satisfy the conditions given: these are orthogonal to $v_i,v_j$ and have scalar product equal to $1$ with $v_k$