# Question about statement of Rank Theorem in Rudin

Theorem Suppose $m,n,r$ are nonnegative integers, $m\ge r, n\ge r$, $F$ is a $C^1$ mapping of an open set $E\subset \mathbb{R}^n$ into $\mathbb{R}^m$, and $F'(x)$ has rank $r$ for every $x\in E$. Fix $a\in E$, put $A = F'(a)$, let $Y_1$ be the range of $A$, and let $P$ be a projection in $\mathbb{R}^m$ whose range is $Y_1$ and let $Y_2$ be the kernel of $P$. Then there are open sets $U$ and $V$ in $\mathbb{R}^n$, with $a\in U\subset E$, and there is a 1-1 $C^1$ mapping $H$ of $V$ onto $U$ (whose inverse is also of $C^1$) such that $$F(H(x)) = Ax+\phi(Ax)\;\;\;\;(x\in V)$$where $\phi$ is a $C^1$ mapping of the open set $A(V)\subset Y_1$ into $Y_2$.

Two questions: 1. Why is $A(V)$ a open set? 2. What does this theorem really say? I lost intuition due to its complicated formulation.

• $A(V)$ is open as a subset of $Y_1$, not (in general) as a subset of $\mathbb{R}^m$. Commented Feb 10, 2014 at 5:20
• A bit late to the party, but I used to be befuddled by this theorem on Rudin too. I later realized that this theorem could be better understood when one is learning manifolds (one good reference would be Lee's smooth manifolds). Commented Sep 30, 2022 at 3:32

Yeah, I think this is a terrible way of looking at the rank theorem. A good way to understand the rank theorem is to consider its consequences, i.e. the inverse and implicit function theorems. I'll try to give a better explanation below. To answer 1, $A(V)$ is open because $V$ is open and $A$ is linear, and non-degenerate linear maps are open. For instance, in $d = 1$ and $A \subset \mathbb{R}$, this boils down to $$A \text{ open } \Rightarrow \{cx : x \in A\} \text{ open }, c \neq 0.$$ Hopefully if you're looking at the rank theorem you know how to generalise this to vector spaces and norms...

As for what the rank theorem is actually saying, let me ask you this question: what is the dimension of the sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? Intuitively, it is 2. Why? Well, save the point at infinity, I have a bijection between points on a plane and points on the sphere (stereographic projection), and a plane is two-dimensional. The rank theorem allows you to say the dimension of the sphere is 2 without knowing about this map. It characterises the dimension of the surface in terms of the rank of the tangent plane at each point. The tangent plane is spanned by 2 linearly independent vectors at every point of the sphere.

You can think about this in many ways -- here are two. Let $E = (0,\pi) \times (0, 2\pi)$ and $F(\phi,\theta) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$. If you compute $F'$, you'll notice that the rank of $F$ is 2 at every point (save degeneracies), so by the rank theorem, for any point on the sphere there is a small patch around it (a neighbourhood) and a $C^1$ map which takes that curved patch to a flat patch $\subset \mathbb{R}^2$.

Another way to do this is to set $$G(x,y,z) = x^2 + y^2 + z^2,$$ $G: \mathbb{R}^3 \rightarrow \mathbb{R}^1$ so that $\nabla G = (2x,2y,2z)$. The rank is 1 for $(x,y,z) \neq 0$, and $\nabla G$ is the normal to the tangent plane, which by the rank theorem always has rank 1. Thus the tangent plane to the sphere always has rank $3 - 1 = 2$ (the 'codimension' of the normal vector in $\mathbb{R}^3$).

A good thinking point is to go back like 150 years. People were really curious about this question: if I have a surface described by a level set, $$F(x,y,z) = 0,$$ can I always write it explicitly as a graph, $$z = f(x,y)?$$ In the case of a sphere, sure: $$x^2 + y^2 + z^2 = 1 \Rightarrow z = \pm \sqrt{x^2 + y^2},$$ but you can certainly cook up examples where this is a lot more difficult (try cubics). The implicit function theorem, a consequence of the rank theorem, answers this question in the affirmative, contingent on properties of the rank of the derivative. This makes sense if you think about the one-dimensional case: you can invert a function $y = f(x)$ at a point $x_0$ if $f'(x_0) \neq 0$ -- in this case the rank is 1 and the dimension is 1, but it's the rank theorem (or inverse function theorem) at work.

EDIT: As a way of remembering the statement, just think about linear maps. If $$A = \begin{pmatrix} \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{pmatrix},$$ $A : \mathbb{R}^4 \rightarrow \mathbb{R}^2$, and you can get at most two dimensions back -- the other two dimensions must get killed. This is the case $m = 2$, $n = 4$, and the rank can be at most $2$. The image can be the whole space. If $$A = \begin{pmatrix} \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{pmatrix},$$ $A : \mathbb{R}^3 \rightarrow \mathbb{R}^4$ and you can get all your dimensions back, but you can't span the whole target space.

• What do you mean by "non-degenerate linear maps"? I know that $A$ has rank $r$ that's it.
– mez
Commented Feb 9, 2014 at 20:04
• Non degenerate: $r \neq 0$, i.e. $A \neq 0$.
– snar
Commented Feb 9, 2014 at 20:07
• Excuse me but Dimension is defined for a linear space, isn't it? A sphere is not a linear space and there is no way for you to find a basis consisting of two vectors such that all the other vector on the sphere is the linear combination of the two. Commented Jul 23, 2018 at 20:22
• What is a vector on the sphere $S^2 = \{(x, y, z) : x^2 + y^2 + z^2 = 1\}$? Do you mean a vector $v_p \in \mathbb{R}^3$ which is tangent to the sphere at a given point $p = (x, y, z)$? If so, and you mean that $v_p$ need not be tangent to the sphere at some other point $p' = (x', y', z')$, you are correct. This is why we attach to /each/ point $p\in S^2$ a vector space $T_p$, which can be thought of as the tangent plane at the point $p$. The vector space $T_p$ is two-dimensional at every point $p \in S^2$. This reflects the fact that the sphere is two-dimensional.
– snar
Commented Jul 24, 2018 at 16:09
• @DanielLi The dimension here is a generalized version of what you've said. It's more like the dimension of a manifold (i.e. have a small open neighborhood around each point that is homeomorphic to R^2). Alternatively, you can think of it as the dimension of the tangent space. Commented Sep 30, 2022 at 3:26