Let V be a vector space Let U and W be subspaces of V Prove

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.

Solution

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 .