Algorithms for quantifying intersections of subspaces

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SUMMARY

The discussion focuses on methods for finding a basis for the intersection of two subspaces V and W within a vector space U. The user highlights the inefficiency of using the identity V ∩ W = (V⊥ ∪ W⊥)⊥, which requires an inner product. Instead, they explore alternative approaches using linear transformations, specifically representing subspaces as either the image or kernel of these transformations. The algebraic method discussed involves solving the equation [A | -B] [x; y] = 0 to find the intersection.

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Greetings,

I'd like to know how one goes about finding a basis for the intersection of two subspaces V and W of a given vector space U. I am aware of the identity [tex]V \cap W = (V^{\per} \cup W^{\per})^{\per}[/tex] (essentially the orthogonal space of the union of orthogonal spaces of V and W), but this isn't computationally efficient. Furthermore it requires an inner product to be defined on U. Are there any other methods of computing a basis for [tex]V \cap W[/tex] if given bases for both V and W (methods that may not require conditions such as the one mentioned above)?

EDIT: My perpendicular symbols didn't show.
 
Last edited:
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There are two good ways of representing a subspace

(1) As the image of a linear transformation
(2) As the kernel of a linear transformation

(Note that a basis is just (1) in disguise; put your basis vectors in as the columns of a matrix)

Either way, you just work through the algebra. If you used (1) for both of your subspaces, then (if V is m dimensional, and W is n dimensional)

V = {Ax | x in R^m}
W = {By | y in R^n}

Then, you're looking for the set
[itex]V \cap W[/itex] = {z : there exists x and y such that Ax = By = z}
= {Ax : x in R^m and there exists y such that Ax = By}



Oh, just in case it's not clear, Ax = By if and only if:

[tex] \left[<br /> \begin{array}{c|c} A & -B \end{array}<br /> \right] \left[ \begin{array}{c} x \\\hline y \end{array}<br /> \right] = 0[/tex]
 
Last edited:
I see. That makes perfect sense, Hurkyl. Thank you so much.
 

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