Let U and V be subspaces of avectoe space W, ans suppose that U∩V is the trivial (zero) space of W.?

Question

(under these conditions, the subspaces U+V = {u+v⎢u ∈ U, v ∈ V} of W is called the internal direct sum of Uand V.) let S be a linearlly independent set of finitely many vectors in U such that U is the span os S, and let T be a linearlly indpendent set of finitely many vectors in V such that V span T. Show that U+V is the span of S∪T and that S∪T is linearlly independent of W.

Answer 1

ok, by the way you have defined S, S is a basis for U. similarly, T is a basis for V. and we need to prove that SUT is a basis for U+V in W.



it is easy to see that span{SUT} = U+V. if u+v is any arbitrary vector in U+V, then:



u is in the span of S, and v is in the span of T, so



u = a1s1 + a2s2 +....amsm, v = b1t1 + b2t2 +...+bntn



where m = dim U = |S|, n = dim V = |T|.



thus u+v = a1s1 + a2s2 +...+ amsm + b1t1 + b2t2 + ...+ bntn



so U+V is contained in span{SUT}. but if w is in span{SUT}, then



w = c1s1 + c2s2 +....+ cmsm + c(m+1)t1 + c(m+2)t2 +....+ c(m+n)tn



and c1s1 + c2s2 +....+ cmsm is in U, c(m+1)t1 + c(m+2)t2 +....+ c(m+n)tn is in V,



so w is in U+V. hence span{SUT} = U+V.



suppose now that:



c1s1 + c2s2 +....+ cmsm + c(m+1)t1 + c(m+2)t2 +....+ c(m+n)tn = 0



this is of the form u + v for c1s1 + c2s2 +....+ cmsm in U, and



c(m+1)t1 + c(m+2)t2 +....+ c(m+n)tn in V, and so is in U+V.



but u + v = 0 implies v = -u, which implies v is in U, so v is in U∩V, so v = 0.



this, in turn implies u = 0. hence:



c1s1 + c2s2 +....+ cmsm = 0 and c(m+1)t1 + c(m+2)t2 +....+ c(m+n)tn = 0



by the linear independence of S, c1 = c2 = ...=cm = 0.



by the linear independence of T c(m+1) = c(m=2) = ....c(m+n) = 0,



so {SUT} is linearly independent in W, and thus forms a basis for the subspace U+V.

Answer 2

The condition of a series being a subspace is such as: a million) The set is closed below vector addition 2) The set is closed below scalar multiplication (by potential of any scalar) 3) The set includes the 0 vector Checking a million), if u and v are in the intersection, then u and v are in the two V and W. The above circumstances carry for V and W, so u + v is in the two V and W. subsequently u + v is likewise in the intersection, and the intersection is closed below vector addition. Checking 2) is fairly comparable. Given u in V and W, and a scalar c, it follows from V and W being subspaces that cu is in the two V and W, and for this reason their intersection. Checking 3) is trivial; 0 lies in the two V and W, and for this reason the intersection.

Answer 3

es U+V = {u+v⎢u

The slightly long way to do this is to note that if u, v and w are roots, than z^3 - 1 = 0 = (z - u)(z - v)(z - w), and expand the right-hand product. If you do that, you'll find that the coefficient of the z term is (uv + vw + uw). In this case, the coefficient of z on the left-hand side is 0, so we can reason that u*v + v*w + u*w = 0, without actually calculating the roots.