Linear Algebra
17.18 min read

Isovectors and the Tangent Space

Let f(x)f(\mathbf{x}) be a differentiable scalar function on Rn\mathbb{R}^n. An isovector of ff at x0\mathbf{x}_0 is a vector u\mathbf{u} along which ff does not change (to first order): Df(x0)u=0Df(\mathbf{x}_0)\mathbf{u} = 0.

The isovectors form exactly the null space of Df(x0)Df(\mathbf{x}_0), a 1×n1 \times n matrix. Assuming Df(x0)≠0Df(\mathbf{x}_0) \neq \mathbf{0}, this null space has dimension n−1n-1 — a hyperplane in Rn\mathbb{R}^n.

An isosurface Sk={x:f(x)=k}S_k = \{\mathbf{x} : f(\mathbf{x}) = k\} is the level set of ff. The isovectors at x0∈Sk\mathbf{x}_0 \in S_k form the tangent space Tx0SkT_{\mathbf{x}_0} S_k — the set of directions you can move at x0\mathbf{x}_0 while staying on SkS_k to first order.

Think of a hiker on a mountain: the isosurfaces are the altitude contours. Moving along a contour means moving in an isovector direction. The gradient points directly uphill, perpendicular to the contours.

Formal View

Definition 17.1 — Isovector
An isovector of ff at x0\mathbf{x}_0 is a vector u\mathbf{u} such that Df(x0)u=0Df(\mathbf{x}_0)\mathbf{u} = 0. The set of isovectors is Null(Df(x0))\text{Null}(Df(\mathbf{x}_0)).
Definition 17.2 — Tangent Space to an Isosurface
Let Sk={x:f(x)=k}S_k = \{\mathbf{x} : f(\mathbf{x}) = k\} be an isosurface and x0∈Sk\mathbf{x}_0 \in S_k with Df(x0)≠0Df(\mathbf{x}_0) \neq \mathbf{0}. The tangent space Tx0SkT_{\mathbf{x}_0} S_k is the space of isovectors:
Tx0Sk=Null(Df(x0))T_{\mathbf{x}_0} S_k = \text{Null}(Df(\mathbf{x}_0))
This is an (n−1)(n-1)-dimensional subspace.

Why This Matters

The tangent space is the correct notion of "directions you can move along a surface." It is the foundation for constrained optimization.

  • Defining velocity vectors of curves on surfaces in differential geometry
  • Constraint Jacobian in robot kinematics describes the tangent space of the constraint surface
  • Tangent spaces appear in Riemannian optimization (optimizing over manifolds)
  • Physics: configuration spaces of mechanical systems are constraint surfaces

Quiz

Question 1

For f(x,y)f(x,y) with Df(x0)=[4,2]Df(\mathbf{x}_0) = [4, 2] (nonzero), what is the dimension of the tangent space Tx0SkT_{\mathbf{x}_0} S_k?

Question 2

An isovector u\mathbf{u} at x0\mathbf{x}_0 satisfies:

Question 3

If x0\mathbf{x}_0 is a critical point of ff, then every vector is an isovector of ff at x0\mathbf{x}_0.

Question 4

For f(x,y)=x2+y2f(x,y) = x^2 + y^2 at x0=(1,0)\mathbf{x}_0 = (1, 0), find a nonzero isovector.

Common Mistakes

  • Confusing isovectors (directions in the domain) with level set points (points in the domain where f=kf = k).
  • Thinking the tangent space is the set of points on the surface — it is a linear subspace of vectors at x0\mathbf{x}_0.
  • Assuming u\mathbf{u} being an isovector means f(x0+u)=f(x0)f(\mathbf{x}_0 + \mathbf{u}) = f(\mathbf{x}_0) exactly — that is only true to first order.