Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $E$ be a finite dimensional real inner product space. I want to define the angle between two subspaces $E_1$ and $E_2$. This has a fairly obvious meaning if $E_1$ is 1-diemsnional: Take the angle between any non-zero vector in $E_1$ and its orthogonal projection onto $E_2$.

There are a number of other cases that can be treated ad-hoc, if one is a hyperplane, or the dihedral angle between planes in $R^3$.

In general, it isn't quite clear what the right definition is. I see two possibilities:

  1. If $p=\dim E_1\le \dim E_2$, consider the two subspace $\lambda^p(E_1)$ and $\Lambda^p(E_2$ of $\Lambda^p(E)$ (which is also an inner product space, and proceed as above, since $\Lambda^p(E_1)$ is a line.

  2. $Hom(E,E)$ is itself an inner product space with the inner product $$ \langle A,B\rangle=trace A^\top B. $$ Let $A_i$ be the orthogonal projection onto $E_i$ and take the angle between $A_1$ and $A_2$.

Are either of these definitions standard? Are they equivalent (I think so)? Is there another definition, perhaps more immediate?

share|improve this question
2  
Didn't you try Google? If you put "angle between subspaces" into Google you will find a ton of stuff there. –  Dick Palais Jul 13 '12 at 17:00
2  
I did go to google. I found lots of things, along the lines of principal angles and the product of their cosines. I don't really understand what that measures; maybe it is one (or both) of the suggestions above. By the way, I have a third possibility: Take the infimum of the angles between pairs of unit vectors, one in $E_1$ and one in $E_2$, and both orthogonal to $$E_1\cap E_2$ (angle $0$ if this set is empy, i.e., if one subspace is a subspace of the other). –  John Hubbard Jul 13 '12 at 17:16
    
John, what are $\lambda^p$ and $\Lambda^p$? –  Vidit Nanda Jul 13 '12 at 19:16
    
I think the two definitions aren't equivalent. If one space is generated by the $p$ first vectors of an ON basis, and $B$ is the matrix of orthogonal proj. on a second same dimension subspace, with obvious 2x2 block partition, then the first angle has cosine $\det(B_{11})$ and the second has cosine $tr(B_{11})/p$. In general, I would say that the most general notion of "angle" is the orbit of the pair of subspaces under the orthogonal group. –  BS. Jul 14 '12 at 14:14
    
I am not sure it can be done in general if $E_1$ and $E_2$ are of different dimensionality. Ideally the angle (or rather its cosine) would be given by the scalar or inner product of $\Lambda_1$ and $\Lambda_2$ where each $\Lambda$ is the exterior product of all the elements in some basis of $E_1$ and $E_2$ respectively, normalised to unity. However the standard definition for the inner product does not apply if $\Lambda_1$ and $\Lambda_2$ are not of the same grade. –  AlexArvanitakis Jul 15 '12 at 1:18

2 Answers 2

There is a standard answer: Principal angles, see http://en.wikipedia.org/wiki/Principal_angles.

Let $p \ge q$ be the dimensions of the two subspaces $E_1$ and $E_2$. Then there is a unique non-increasing sequence $[c_1,c_2,...,c_q]$ with entries in $[0,1]$ (and a matching non-decreasing sequence $[s_1,s_2,...,s_q]$) such that one can have an orthonormal basis for $E$, call it $e_1,e_2,...$, in such a way that one subspace is generated by orthonormal vectors $$e_1,e_2,...,e_p$$ and the other subspace generated by orthonormal vectors $$c_1e_1+s_1e_{p+q},c_2e_2+s_2e_{p+q-1},...,c_qe_q+s_qe_{p+1}.$$ One can see this from the Singular Value Theorem. The principal angles are obviously those angles whose cosines match the $c_i$ values.

This concept captures all of the geometric invariant information relating the positioning of the two subspaces, so any well-defined definition you care to give must be a deterministic function of this sequence of principal angles.

share|improve this answer

Let me confuse you some more. There is a third possibility that is used frequently in functional analysis. Define

$$\delta(E_1,E_2)= \sup_{x\in E_1,\;|x|=1}{\rm dist}\; (x,E_2). $$

The number $\delta(E_1,E_2)$ is called the gap between $E_1$ and $E_2$. Clearly $\delta(E_1, E_2)\in [0,1]$ so that there exists $\theta\in [0,\frac{\pi}{2}]$ such that

$$\delta(E_1,E_2)=\sin \theta.$$

We define the above $\theta$ to be the angle between $E_1,E_2$. Note that if $\dim E_1=1$, than this definition agrees with your first definition. However

$$\delta(E_1, E_2)\neq \delta(E_2,E_1).$$

Moreover

$$ \theta <\frac{\pi}{2} \Longleftrightarrow \delta(E_1,E_2)<1 \Longleftrightarrow E_1\cap E_2^\perp= 0. $$

Your first definition of angle has a similar property. Finally let me point out that

$$ \delta(E_1,E_2)= \Vert P_{E_2^\perp}P_{E_1}\Vert, $$

where $P_U$ denotes the orthogonal projection onto the subspace $U$, and for any linner operator $A$ we set

$$ \Vert A\Vert =\sup_{|x|=1} |Ax|. $$

share|improve this answer
    
If you take the Hausdorff distance between the intersections of $E_1$ and $E_2$ with the unit sphere, you get a very good metric on the space of all subspaces. It works even for closed subspaces of Banach spaces. This is similar to a symmetrized version of your $\delta$. It isn't quite the angle I am after, because it returns $\pi/2$ when $E_1\subset E_2$ and $E_1 \ne E_2$. I would want the angle between a line and a plane in $R^3$ to be $0$ when the line is in the plane. –  John Hubbard Jul 15 '12 at 16:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.