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Is there a usual distance between linear subspaces ($V,W$) of an n-dimensional normed vector space with inner product?

In the case of hyper-planes one could use the angle (based on the inner product of the vector space). What can be used in the case of subspaces with lower dimension (not necessarily equal)? e.g. $dim(V)= n-2$ and $dim(W) = n-4$

Thanks

JuanPi
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    Well, the word "distance" is already taken. If you call something "distance", you want it to be a distance, i.e. a metric. Your question is rather vague, since you do not add any reasonable condition that your "distance" should satisfy. – Siminore Sep 17 '12 at 17:03
  • I need the usual properties of a distance (maybe measure?) d(x,x)=0 iff x=x, d(x,y)>= 0 , etc... – JuanPi Sep 17 '12 at 17:49

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There is, if both subspaces have the same dimension. You can actually make the set of $k$-dimensional subspaces of a vector space $V$ into a metric space (a manifold, in fact) called the Grassmannian, denoted $\mathrm{Gr}(k,V)$. The distance between two subspaces $W$ and $W'$ is then $\|P_W-P_{W'}\|$ where $P_X$ denotes projection onto $X$ and $\|\cdot\|$ is the operator norm.

Alex Becker
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  • Thanks Alex Becker! What happens when the subspaces are of different dimension? – JuanPi Sep 17 '12 at 17:47
  • @JuanPi Then there is no notion I know of. If the dimension of $V$ is $n$ and the dimensions of $W,W'$ are $k,n-k$ we can use the duality of $k$ and $n-k$ dimensional subspaces to consider both as $k$ dimensional subspaces, but otherwise I have no idea. – Alex Becker Sep 17 '12 at 17:55
  • Why not use the same definition regardless of dimension? – Jonas Meyer Sep 18 '12 at 03:48
  • @JonasMeyer How would that let you compare two subspaces of different dimension? The Grassmannian is dimension-specific. – Alex Becker Sep 18 '12 at 03:55
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    @AlexBecker: There is no problem finding distances between orthogonal projections with ranges of arbitrary dimension. The set of subspaces with this distance might be of interest, depending on what JuanPi wants to use it for. It amounts to the same as using your definition on each Grassmannian, and defining the distance between spaces of different dimensions to be $1$. – Jonas Meyer Sep 18 '12 at 04:02
  • @JonasMeyer Ah, I see what you mean. – Alex Becker Sep 18 '12 at 04:22