A tutorial sheet has the following problem.
Find a unit normal vector and a basis for the tangent space of the following smooth manifold $M \subseteq \mathbb{R}^2$ at a point $(a,b) \in M$. $$M=\{(x,y) \in \mathbb{R}^2 : x^2+y^2=1\}$$
The idea is to do it without finding an explicit parametrization for $M$ (although of course, that is quite easy.)
The solution sheet says this:
Have $f(x,y) = x^2+y^2 = 1$ so a normal is $\nabla f = (2x,2y) = (2a,2b)$ at $(x,y) = (a,b)$.
Tangent space = $\mathrm{ker}(Df) = (\nabla f)^\perp = \{v \in \mathbb{R}^2 : v \cdot (2a,2b) = 0\}$ has basis $\{(-b,a)\}$.
The stuff about normals makes sense, but I don't get the line about tangent spaces. In particular:
- What is the difference between $Df$ and $\nabla f$? Aren't they the same?
- How do we get from $\mathrm{ker}(Df)$ to $(\nabla f)^\perp$? What is the general principle here?
1) Yes, $Df$ and $\nabla f$ are pretty much the same. By definition, the derivative $Df_{(a,b)}$ is the best linear approximation to the function $f$ at the point $(a,b)$. It turns out that the matrix of $Df_{(a,b)}$ with respect to the standard basis is the gradient $\nabla f$.
2) This just follows by the definition of matrix multiplication. Given a matrix $A$, if $x \in \ker A$ (here we consider left multiplication by $A$ as a linear transformation), then $A\vec{x} = \vec{0}$. By the definition of matrix multiplication, to compute the $i^\text{th}$ entry of $A\vec{x}$ we take the dot product of the $i^\text{th}$ row of $A$ and the vector $\vec{x}$. Since $A\vec{x} = \vec{0}$, then the dot product of each row of $A$ with $\vec{x}$ is $0$, so $\vec{x}$ is orthogonal to each of the rows of $A$.