Dual Basis Vectors

The creation of a vector space instantly creates dual vector space. When you created the vector space you had to choose a field. You can take any combination of a vector from your vector space and a scalar from your field and whatever it takes to map from the vector to the scalar the “whatever” is found in the Dual Vector space. That whatever is a vector.

You can create any basis set you want and use it to create a dual vector and that vector will dot product with any vector from the vector space to give a scalar.

However, you can make a shrewd choice–you use a calculation for which you input the basis vectors you chose for your vector space and it will output a set of basis vectors to use in the dual vector space.

Certain things we want to have will come true. When the going gets strange, that dual basis vector set that was calculated will help you. You will be on a trek on a path and every infinitesimal step you take brings you to a new tangent space and that brings a change to your calculation.

  • k=i \rightarrow \delta^k _i=1
  • k\neq i \rightarrow \delta^k _i=0

\begin{bmatrix} \delta^1_1=1 && \delta^1_2=0 && \delta^1_3=0 \\ \delta^2_1=0 && \delta^2_2=1 && \delta^2_3=0 \\ \delta^3_1=0 && \delta^3_2=0 && \delta^3_3=1 \end{bmatrix}

And it might help to just see the answers:

\begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix}

Now that you know the answers, we can reverse engineer that you will vectors:

e^1 =(x,y)

e_1 = (1,0)

e^1 \cdot e_1 = (x)(1) + (y)(0) = 1

e^1 \cdot e_1 = x + 0 = 1

After ‘x’ is forced to become one, ‘y’ has no choice but to become zero.

e^1 =(x,y)

e^1 =(1,0)

One should not conclude that basis vectors into a basis vectors are the same. Consider the following pairing:

  • e^1 = (2,0)
  • e_1 = (\frac{1}{2},0)

It gets even more interesting if we are working with complex numbers. It will be necessary to do a sign flip so that the multiplication well cancel out the two eyes two negative one and that sign flip will counter the negative of the negative one.

  • (+1i)*(+1i) = -1
  • (+1i)*(-1i) = -(-1) = 1