Is the Intersection of T-Invariant Subspaces Always T-Invariant?

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SUMMARY

The intersection of any collection of T-invariant subspaces of a vector space V is itself a T-invariant subspace. The proof requires demonstrating that for any collection of T-invariant subspaces Wα, if v belongs to the intersection, then T(v) also belongs to the intersection. The discussion highlights that the initial proof incorrectly limited the argument to only two subspaces, W1 and W2, rather than addressing an arbitrary collection, which is essential for completeness. The correct approach involves using an index set to represent the subspaces.

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Homework Statement


Prove that the intersection of any collection of T-invariant subspaces of V is a T-invariant subspace of V.

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The Attempt at a Solution


Let W1 and W2 be T-invariant subspaces of V. Let W be their intersection.
If v\inW, then v\inW1 and v\inW2. Since v\inW1, T(v)\inW1 & v\inW2, T(v)\inW2. Therefore T(v)\inW.
For any x,y\inW, x,y\inW1 and x,y\inW2, x,y\inW and cx+y\inW since W1 and W2 are subspaces.
Thus the intersection of any collection of T-invariant subspaces is a T-invariant subspace.

My answer was marked wrong. The grader's comment was to cross out "any" and replace it with "2". What I should have said? Was I supposed to explicitly point out that this applies from 2...n? :(
 
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Your grader is right. Your proof shows that given two invariant spaces W1 and W2, that their intersection is invariant.
Your proof does not shows that given n invariant subspaces W1, W2,..., Wn, that their intersection is invariant.
 
In fact, the problem says "the intersection of any collection" so showing this for "n invariant subspaces" would not be sufficient. The collection does not have to be finite nor even countable.

Your proof will work nicely if you modify it so that instead of "W1" and "W2", you use W<sub>\alpha</sub> where \alpha is simply some label distinguishing the different sets- that is, \alpha is just from some index set, not necessarily {1, 2}, nor even a set of numbers.
 

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