Basis and Duality

  • We have seen that vectors live in spaces with a certain number of dimensions. But what defines those dimensions? This is where basis vectors come in.

  • A basis is a set of vectors that can build every other vector in the space through scaling and adding (linear combination), with no redundancy. They are the building blocks of the space.

  • A basis must satisfy two conditions:

    • Linearly independent: No basis vector can be built from the others. Each one contributes a genuinely new direction.

    • Spanning: Every vector in the space can be expressed as a combination of the basis vectors. Nothing is left out.

  • The number of vectors in a basis equals the dimension of the space. In $\mathbb{R}^2$ you need 2, in $\mathbb{R}^3$ you need 3, and so on.

  • The most natural basis is the standard basis, the unit vectors along each axis:

    • In $\mathbb{R}^2$: $\hat{\mathbf{i}} = (1, 0)$ and $\hat{\mathbf{j}} = (0, 1)$
    • In $\mathbb{R}^3$: $\hat{\mathbf{i}} = (1, 0, 0)$, $\hat{\mathbf{j}} = (0, 1, 0)$, $\hat{\mathbf{k}} = (0, 0, 1)$

  • Any vector is just a weighted sum of these basis vectors. The vector $(3, 2)$ is really $3\hat{\mathbf{i}} + 2\hat{\mathbf{j}}$. The weights (3 and 2) are the coordinates of the vector in that basis.

  • But the standard basis is not the only valid basis. In $\mathbb{R}^2$, the vectors $(1, 1)$ and $(-1, 1)$ also form a basis. They are linearly independent and can reach any point in the plane. The same vector will just have different coordinates in this new basis.

  • A change of basis re-expresses the same vector using different building blocks. The vector has not moved, we are just describing it from a different perspective.

  • This is done by multiplying by a change of basis matrix $P$, whose columns are the new basis vectors written in the old coordinates. To go back, multiply by $P^{-1}$.

  • In ML, change of basis appears frequently. PCA, for example, finds a new basis (the principal components) where the data is easier to understand, the axes align with the directions of greatest variation.

  • Now, there is a deeper idea hiding here. When we write $\mathbf{v} = (3, 2)$, the coordinates 3 and 2 are really the result of "measuring" $\mathbf{v}$ along each basis direction. The first coordinate asks "how much of $\hat{\mathbf{i}}$ is in $\mathbf{v}$?", the second asks "how much of $\hat{\mathbf{j}}$?"

  • Each of these measurements is a linear functional, a function that takes a vector and returns a single number. The collection of all such linear functionals forms the dual space $V^\ast$.

  • Think of it this way: vectors are the objects, and linear functionals are the rulers that measure them. The dual space is the set of all possible rulers.

  • For every basis ${\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n}$, there is a corresponding dual basis ${\mathbf{e}_1^\ast, \mathbf{e}_2^\ast, \ldots, \mathbf{e}_n^\ast}$. Each dual basis vector extracts exactly one coordinate:

$$ \mathbf{e}_i^\ast(\mathbf{e}_j) = \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases} $$

  • $\mathbf{e}_1^\ast$ returns 1 when applied to $\mathbf{e}_1$ and 0 for everything else. It perfectly isolates the first coordinate.

  • The dot product connects these two worlds. When you compute $\mathbf{u} \cdot \mathbf{v}$, you can think of one vector acting as a "ruler" measuring the other. The dot product $\mathbf{u} \cdot \mathbf{v}$ is the same as applying the linear functional defined by $\mathbf{u}$ to the vector $\mathbf{v}$.

  • This means every vector secretly defines a linear functional, and every linear functional can be represented by a vector. In finite dimensions, the dual space is essentially a mirror image of the original space.

  • Duality may seem abstract now, but it underlies many practical ideas: coordinates are dual basis evaluations, the dot product is a duality pairing, and transformations like attention in neural networks operate by having one set of vectors "query" another, which is duality in action.

Coding Tasks (use CoLab or notebook)

  1. Express a vector in two different bases and verify they represent the same point. Try creating your own basis and see what coordinates the vector gets.
import jax.numpy as jnp

v = jnp.array([3.0, 2.0])

# Standard basis: coordinates are just the components
print(f"Standard basis coords: {v}")

# New basis: (1,1) and (-1,1)
P = jnp.array([[1.0, -1.0],
               [1.0,  1.0]])
new_coords = jnp.linalg.solve(P, v)
print(f"New basis coords: {new_coords}")

# Verify: reconstruct from new coords
reconstructed = new_coords[0] * P[:, 0] + new_coords[1] * P[:, 1]
print(f"Reconstructed: {reconstructed}")
  1. Verify the dual basis property: each dual basis vector extracts exactly one coordinate and returns zero for the others.
import jax.numpy as jnp

# Standard basis in R3
e1 = jnp.array([1.0, 0.0, 0.0])
e2 = jnp.array([0.0, 1.0, 0.0])
e3 = jnp.array([0.0, 0.0, 1.0])

v = jnp.array([5.0, 3.0, 7.0])

# Each dot product extracts one coordinate
print(f"e1 · v = {jnp.dot(e1, v)}")
print(f"e2 · v = {jnp.dot(e2, v)}")
print(f"e3 · v = {jnp.dot(e3, v)}")