# How many parameters are required to specify a linear subspace?

A problem in Peter Lax's Linear Algebra involves looking at the family of $n\times n$ self-adjoint complex matrices and asking: on how many real parameters does the choice of such a matrix depend? This made me ask: on how many real parameters does a choice of a $k$-dimensional subspace of $\mathbb{R}^n$ depend? Clearly the $n-1$-dimensional subspaces as well as the $1$-dimensional subspaces are an $n-1$ parameter family, since we can specify a $1$-dimension subspace by choosing a vector up to scaling, and we can choose an $n-1$-dimensional subspace by choosing its one-dimensional orthogonal complement. This suggests that the formula should be something like ${n\choose k}-1$. I'm having trouble proving this...any ideas?

• I would rather say that for a $k$-dimensional subspace $V\subseteq \Bbb R^n$, you need $k(n-1)$ parameters to describe it directly (a basis of vectors up to scaling), and $(n-k)(n-1)$ to describe indirectly. Though I have no proof at the moment that this is indeed the sharpest bounds possible. – Arthur Jun 21 '13 at 7:44
• This may be helpful: en.wikipedia.org/wiki/Grassmannian – user64480 Jun 21 '13 at 7:52
• @user64480 Indeed, after some scrolling you'll find there: "In particular, the dimension of the Grassmannian is r(n–r);." – Elmar Zander Jun 21 '13 at 17:17

You can generate an arbitrary linear subspace of dimension $d$ by applying an orthogonal transformation on the subspace spanned by the first $d$ basis vectors. However, orthogonal transformations which happen completely inside the subspace have no effect, nor do orthogonal transformations which happen completely in its complement.

Since you need $k(k-1)/2$ parameters to specify an orthogonal transformation in $k$ dimensions, the number of arguments to specify a $d$-dimensional linear subspace of $\mathbb R^n$ is $$\frac{n(n-1)}{2} - \frac{d(d-1)}{2} - \frac{(n-d)(n-d-1)}{2}$$

To specify an affine subspace of dimension $d$, you need $n-d$ additional parameters to describe the displacement from the origin (displacements in the subspace don't change it).

Here are some examples:

To specify a $0$-dimensional linear subspace, you need $n(n-1)/2 - 0 - n(n-1)/2 = 0$ parameters. That's obvious, because there's only one 0-dimensional linear subspace.

To specify a $1$-dimensional linear subspace (a straight line through the origin), you need $n(n-1)/2 - 0 -(n-1)(n-2)/2 = n-1$ parameters. This is also clear; to specify a one-dimensional subspace, you need to specify one point on the unit sphere, which is $n-1$-dimensional.

To specify a $2$-dimensional linear subspace (a plane through the origin) in $\mathbb R^3$, you need $3\cdot2/2 - 2\cdot1/2 - 1\cdot0/2 = 2$ parameters.

More generically, from the formula you see that you need the same number of parameters to specify a linear subspace and to specify its complement, which is also obvious because one implies the other.

To specify a $1$-dimensional affine subspace (a general straight line) in $\mathbb R^n$, you need $n(n-1)/2 - 0 - (n-1)(n-2)/2 + (n-1) = 2(n-1)$ parameters. This can also be seen directly from the fact that you can write the points of the line as $\mathbf u + \lambda\mathbf v$, (2 vectors $\implies$ 2n parameters) but the length of $v$ is irrelevant (1 parameter less), and $\mathbf u$ can be changed by an arbitrary multiple of $\mathbf v$ (again, 1 parameter less).

Edit: As Elmar Zander noted, the formula for linear subspaces can be simplified significantly to $d(n-d)$.

Obviously this means that the number of affine subspaces then also simplifies to $(d+1)(n-d)$.

• Fantastic answer, thanks! – Eric Auld Jun 21 '13 at 9:23
• Great insight! I just wanted to added that the formula simplifies significantly to $d(n-d)$. You can see that even without calculation, because you have triangular numbers, and from the big triangle of edge length $n$ you're subtracting smaller triangles of edge lengths $d$ and $n-d$, such that you're left with a rectangle with side length $d$ and $n-d$. – Elmar Zander Jun 21 '13 at 10:20
• @Elmar, great simplification and insight. One thing I didn't understand: isn't $n(n-1)/2$ a triangle of edge length $n-1$, not $n$? – Eric Auld Jun 22 '13 at 12:03
• @ElmarZander: Thanks to noting this. I'll add this to the answer. – celtschk Jun 22 '13 at 15:05
• @EricAuld. You're right, the triangles have edge lengths (or rather: number of dots on each side) of $n-1$, $d-1$ and $n-d-1$ respectively. I was a bit sloppy here, but the result is the same, as you can easily visualize. – Elmar Zander Jun 22 '13 at 23:26