Algebra?
You have problems with this proof: Let V be a vector space over an infinite field. Show that V is not the Union, many good fineitely Subspaces of V. I'm assuming that the union of V is a finite number of proper subspaces of Wi where i 'is only a few = 1 <= n and n is minimal. This means we can assume that W1 is not contained in the rest of the W2 unions … . Wn Let V in V W1 W2 not in the union … . Wn V and not W W1. Since W1 is an eigenspace. Consider A = (v + w in F), an infinite set of vectors. So thats what I expected at the start but I do not know where there are elements of S … May I begin? Could someone add something? Thank you!
rosurcep fact, an error says: He jump on the money that does not return the Union to your problem: a better account of the set A = (w + AV | a in F). Now A is an infinite set (why?) The elements of A are not in W1 (why?) So they have to in W2 are … Wn. One of these subspaces, say without loss of generality, W2 contain two different elements of A: W + + AV A'v W. This means that their difference (AA ') v is in W2, contradiction.
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