Let V be a vector space. Let U and W be subspaces of V. Prove: U intersection W
ID: 3027565 • Letter: L
Question
Let V be a vector space. Let U and W be subspaces of V. Prove: U intersection W is a subspace of V. Prove: U union W is a subspace of V if and only if U subsetorequalto W or W subsetorequalto U. Let U + W: = {v = u + w element V: u element U and w element W}. Prove that U + W is a subspace of V. Prove: Suppose U intersection W = {0}. Suppose that B is a finite basis for U and B_1 is a finite basis for W. Then B union B_1 is a finite basis for U + W. (It is then common to denote U + W as U W, and call it the direct sum of U and W.) Prove: Suppose U + W is finite-dimensional. It is possible to choose a basis B for U and a basis B_1 for W such that B intersection B_1 is a basis for U intersection W. Show by example that, in general, if B is a basis for U and B_1 is a basis for W, then B intersection B_1 is NOT a basis for U intersection W.Explanation / Answer
a]
Let u, v U W and K.
Then u + v U (because U is a subspace) and u + v W(because W is a subspace).
Hence u + v U W.
Similarly, we get v U W,
so U W is a subspace of V.
b]
() This is the easy direction.
If UW or WU, then we have UW=W or UW=U respectively. So UW is a subspace as U and W are subspaces.
() This is the harder direction and I give a direct proof.
Assuming WU, I'll show UW. Let xU and yWU.
So, by the definition of the union, we have xUW and yUW.
Therefore, as UW is a subspace, x+yUW which, again by the definition of the union, means that x+yU or x+yW.
If x+yU then, as U is a subspace, y=(x+y)+(x)U which is impossible as yWU.
So it must be that x+yW in which case, as W is a subspace, x=(x+y)+(y)W.
Therefore, as x was arbitrary, UW as desired.
c]
Let v1,v2 U + W.
Then v1 = u1 + w1 for some u1 U, w1 W and v2 = u2+w2 for some v1 U, v2 W.
Then v1+v2 = (u1+w1)+(u2+w2) U+W.
Similarly, if K then v = u + w U + W.
Thus U + W is a subspace of V .
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