What is the number $N^d_k$ of real-valued parameters that are needed to specify a k-dimensional subspace of $\mathbb{R}^d$? And how can these parameters be interpreted?

I know: $N^d_1 = N^d_{n-1} = d - 1 = \binom{d}{1} - 1$.

The parameters can be interpreted as the d components of a vector spanning the 1-dimensional subspace minus its (arbitrary) length.

I know: $N_1^3 = N^3_2 = 2 = \binom{3}{2} - 1$.

The parameters can be interpreted as two angles or as the three components of a normal vector of the 2-dimensional subspace minus its (arbitrary) length.

I know: $N^d_2 = N^d_{d-2} = \binom{d}{2} - 1$

I believe this, because a d-dimensional rotation has $\binom{d}{2}$ degrees of freedom, one for the rotation angle, the remaining $\binom{d}{2} -1$ ones for the (d-2)-dimensional (hyper)plane of rotation which also defines a 2-dimensional hyperplane as its orthogonal complement.

Question: How do I know that $\binom{d}{2}$ is the number of degrees of freedom of a d-dimensional rotation?

How can these $\binom{d}{2}$ parameters of a rotation or the $\binom{d}{2} - 1$ parameters of a 2-dimensional hyperplane be interpreted (maybe even intuitively)?

I guess that $N^d_k$, the number of parameters that are needed to specify a k-dimensional subspace of $\mathbb{R}^d$, is given by $\binom{d}{k} -1$. How can this be shown? Only formally by mathematical induction or more directly, using e.g. the observation, that there are $\binom{d}{k}$ k-dimensional subspaces of $\mathbb{R}^d$ spanned by k of d elements of an orthonormal basis of $\mathbb{R}^d$?

  • $\begingroup$ As Joel points out below, you have to be really careful about using notions like "how many parameters..." I interpret this question as "What is the dimension of the space of..." which is a more precise way of saying things. $\endgroup$ – Ben Webster Dec 31 '09 at 19:12
  • $\begingroup$ At Ben's suggestion, I have turned my answer into a comment. There is a continuous map from R onto R^d for any d, by interleaving digits, and in this sense just one parameter suffices to specify a subspace, providing a continuous map from R onto the subspaces. I take this to show that one needs to be more precise about the sense of parameterization that is allowed or desired. $\endgroup$ – Joel David Hamkins Dec 31 '09 at 19:21
  • $\begingroup$ Could you please help me to clearify my concept of a "parameter" to make my question sensible. My concept of a "parameter" is something like a "degree of freedom" and is such that any 1-dimensional subspace has d-1 "degrees of freedom" (and corresponding to it its (d-1)-dimensional orthogonal complement). In Joel's interpretation my question patently does not make sense. Nor in Ben's interpretation: I am obviously not looking for the dimension of a k-dimensional subspace. $\endgroup$ – Hans-Peter Stricker Dec 31 '09 at 20:45
  • $\begingroup$ If you mean degree of freedom, then the k(n-k) count is the correct one. What you're generally doing is constructing some space X which aggregates all of the objects you are interested in (in this case, subspaces of a given dimension). If you are lucky, X is "nice" in some ways: maybe homogeneous, maybe has a smooth structure, etc. If X is sufficiently nice, it has the same dimension at each point. This dimension is a rigorous version of "degrees of freedom". $\endgroup$ – Matt Noonan Dec 31 '09 at 20:55
  • $\begingroup$ Thanks for taking my objecction seriously. Ben was suggesting that one look at the dimension, not of the individual subspaces, which are all k as you mention, but rather, the dimension of the collection of all subspaces, as a space of its own. How are we to think about this dimension? I suppose there is some kind of manifold structure here... $\endgroup$ – Joel David Hamkins Dec 31 '09 at 21:40

This answer is similar to what Ben said about $d \times k$-matrices, but maybe a little more visual. Instead of computing the dimension "as a whole", I'll just compute the dimension of subspaces neighboring a given one.

Take some fixed $k$-dimensional subspace $P$ with complementary space $P^\perp$ of dimension $n-k$. Any sufficiently nearby subspace $P'$ to $P$ looks like a graph of a linear function $A : P \to P^\perp$. And on the other hand, any such linear function defines a unique subspace. So you only need to count linear maps from $P$ to $P^\perp$, which are $k(n-k)$-dimensional.

Thinking about the issue locally also helps avoid a mistaken dimension count like $n \choose k$. The problem with the $n \choose k$ count is that you are enumerating some random points on the Grassmannian, which doesn't tell you anything about the dimension. For example, take the case of 2-dimensional subspaces of $\mathbb{R}^3$. A subspace here is determined by its normal vector, so there is a bijection between 2-dimensional (oriented) subspaces and the unit sphere. The $3 \choose 2$ planes in your count correspond to the north pole and two equatorial points $90^\circ$ apart. But you wouldn't conclude that the sphere is 3-dimensional just because it has three points!


The answer to this question is $k(d-k)$ "parameters" actually. One way to make this formal is to say that the Grassmannian of k planes in $\mathbb{R}^d$ is a $k(d-k)$ dimensional manifold.

One way to think about this result is that generically, such a plane is the image a unique $d\times k$-matrix where the first $k\times k$ box is the identity matrix, leaving $k(d-k)$ entries you can choose freely.

You may want to look up some information on "Grassmannians" since there is much more detailed stuff out there on this question.

  • 4
    $\begingroup$ Oddly enough, the wikipedia page on Grassmannian doesn't mention the dimension explicitly, as far as I can see. $\endgroup$ – Reid Barton Dec 31 '09 at 18:22
  • $\begingroup$ It gets very close to it at several points -- take a look at the example in the Schubert cell section where the euler characteristic of grassmanians is computed. $\endgroup$ – Ryan Budney Dec 31 '09 at 19:08

Let me expand a little on Ben's matrix remark, since all this talk about Grassmannians might give you the impression that the answer is complicated! Any $d \times d$ matrix has a rowspace of dimension $k$, the row rank of the matrix. You learned in linear algebra that

  • row operations leave the rowspace of a matrix invariant, and
  • up to row operations, every matrix has a unique representative in reduced row echelon form.

I claim that this allows us to identify $k$-dimensional subspaces of $\mathbb{R}^d$ with rrefs with $k$ nonzero rows. Why? Because any two bases of a given space, when written down in matrix form, are related by row operations.

Now the rrefs with $k$ nonzero rows are exactly as Ben describes them: the $k \times k$ box in the upper left is the identity matrix and the $k(d-k)$ entries to the right of those can be chosen freely. (The fancy term for this is called the "Schubert cell decomposition" of the Grassmannian.)


Here is a more geometric way to find the answer. We can specify a $k$ dimensional subspace by giving $k$ vectors in that subspace. That would require $kn$ parameters, but we are overspecifying. By how much are we overspecifying? Write those vectors in a basis such that the first $k$ coordinates specify the location in the subspace, and the remaining $n-k$ coordinates specify the location orthogonal to the subspace. For specifying a subspace the $k$ coordinates within the subspace do not matter, since whatever values for those coordinates we choose, we get the same subspace. Therefore only the $n-k$ components of those $k$ vectors matter, so that's a total of $k(n-k)$ parameters.

In my opinion such a geometric argument gives far more insight than looking at reduced row echelon form matrices.


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