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direct sum description

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Working note by Phil, dated 1.11.15, with a 1.15.16 remark that it is superseded by his direct sum Appendix B and should be filed in obs. It defines the direct sum by rules on pairs of vectors and checks commutativity, associativity, zero, inverses, scalar identity and distributivity. It ends with an R2 and R3 example giving R5. Symbols are partly lost in extraction, and the draft ends unfinished with some repeated passages.

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Direct sum rewrite PhL 1.11.15 1.15.16: I think this is superseded by my direct sum Appendix B. So file in obs. Brief Digression: The Direct Sum of Vector Spaces Let vi ϵ V and wi ϵ W where V and W are vector spaces, and α ϵ K is a scalar. The direct sum operator can be defined by these rules (axioms), v1w1 + v2w2 + ... + vkwk = (v1+v2 + ...+ vk) (w1+w2 + ... + wk) (1) (αv1)(αv2) .... (αvk) = α(v1v2 .... vk) (2) For k = 2 these say v1w1 + v2w2 = (v1+v2) (w1+w2) (3) Define space Z ≡ VW and let z1 ≡ v1w1 and z2 ≡ v2w2 where z1,z2 ϵ Z. Rule (3) shows that z1 + z2 = z2 + z1 so addition in Z is commutative. Define z3 ≡ v3w3 . Then rule (1) for k = 3 says z1+ z2 + z3 = (z1+ z2) + z3 = z1+ (z2 + z3) so addition is associative in Z. The zero element in Z is 00 since vw + 00 = (v+0)(w+0) = vw The inverse of z = vw is -z = (-v)(-w) since then z + (-z) = vw + (-v)(-w) = (v-v)(w-w) = 00 = 0. For scalars a,b we have a(bz) = (ab)z since a(bz) = a(b[vw]) = a[ (bv)(bw) ] = (abv)(abw) = (ab)(vw) = (ab)z Identity for scalar multiplication is 1(z) = 1(vw) = (1v)(1w) = vw = z Distributive requirement #1 a(z1+z2) = az3 = a(v3w3) = (av3)(aw3) = (av1+av2)(aw1+aw2) = (av1)(aw1) + (av2)(aw2) = a(v1w1) + a(v2w2) = az1+ az2 Distributive requirement #2 (a+b)z = (a+b)(vw) = [(a+b)v][(a+b)w] = [av+bv][aw+bw] = (av)(aw) + (bv)(bw) = a(vw) + b(vw) = az + bz So I am only missing a proof of one vector space requirement! a(z1+z2) = az3 = a(v3w3) = (av3)(aw3) = (av1+av2)(aw1+aw2) = (av1)(aw1) + (av2)(aw2) = a(v1w1) + a(v2w2) = az1+ az2 It is also associative (z1 + z2) + z3 = z1 + (z2 + z3)= If we define z1 ≡ v1w1 and z2 ≡ v2w2, the first rule allows that z1 + z2 = z2 + z1 The object v1w1 is an element of the space Z = VW. In order for Z to be a vector space, I think I have to show that vw is commutative! If we define z1 ≡ v1w1 and z2 ≡ v2w2, the first rule above says z1 + z2 = z2 + z1 not what I want! If for example we let V = R2 and W = R3, one can visualize the direct sum z = vw in this manner v = ϵ V2 w = ϵ V3 z = vw = = ϵ V5 . (5.4.1)