Appendix B

Appendix B This was installed on 1.26.16, do not edit here. Appendix B: Direct Sum of Vector Spaces There are eight short numbered sections below. Here are the eight headings: 1. Axioms for 2. Direct Sum Space VW 3. Basis for VW 4. Z = VW is a vector space 5. vw does not commute 6. Visualization of the Direct Sum 7. Extension to multiple products 8. Application to tensor products 1. Axioms for 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) (B.1) (αv)(αw) = α(vw) . (B.2) In slightly more concise notation (B.1) can be written Σi=1k (viwi) = (Σi=1kvi) (Σi=1kwi). For k = 2 (B.1) becomes, v1w1 + v2w2 = (v1+v2) (w1+w2) . (B.3) Since V and W are vector spaces, each has a 0 element and we can write v10 + 0w2 = (0+v1) (0+w2) = v1w2 // (B.3) with w1 = 0 and v2 = 0 (B.4) (αv)0 = α(v0) // w = 0 0(αw) = α(0w) // v = 0 (B.5) 2. Direct Sum Space VW Define space Z by Z ≡ VW (B.6) and let zi ≡ viwi ϵ Z. (B.7) One might write : (V,W) → VW : (v,w) ↦ (vw) Lemma: Given some zi, we can find vi and wi such that zi = viwi . (B.8) Proof: When we say Z ≡ VW, we mean these spaces are the same, so there is a 1-to-1 correspondence between elements of zi ϵ Z and elements viwi ϵ VW. 3. Basis for VW Let us assume that: ei form a basis of dimension n for V e'j form a basis of dimension n' for W Fact: A basis for Z can be written as {e10, e20 ........en0, 0e'1, 0e'2, ....0e'n'} (B.9) Proof: Let vi be components of vector v, and wi the components of vector w. Consider: { v1(e10) + v2(e20) + ... + vn(en0)} + {w1(0e'1) + w2(0e'2) + ... + wn'(0e'n')} ={ (v1e1)0 + (v2e2)0 + ...+ (vnen)0} + {0(w1e'1) + 0(w2e'2) + ... + 0(wn'e'n')} // (B.5) = { (v1e1+ v2e2 + ...+ vnen )(0+0..+0)} + {(0+0+..+0) (w1e'1+ w2e'2 + ...+ wne'n )} // (B.1) = (v1e1+ v2e2 + ...+ vnen ) 0 +0 (w1e'1+ w2e'2 + ...+ wne'n ) = (v1e1+ v2e2 + ...+ vnen ) (w1e'1+ w2e'2 + ...+ wne'n ) // (B.4) = v w This z = v w is an arbitrary element of Z, and we have therefore shown that an arbitrary element of Z can be expanded on the basis shown in (B.9) and that no smaller basis will do the job. QED Fact: If dim(V) = n and dim(W) = n'. then dim(VW) = n + n' (B.10) Proof: Count the basis elements shown in (B.9). Compare this Fact with that shown in (4.1.1) : Fact: If dim(V) = n and dim(W) = n', then dim(VW) = n * n' . (4.1.1) 4. Z = VW is a vector space Fact: If V and W are vector spaces, then Z = VW is a vector space. (B.11) Proof: We just run down the required axioms listed for example on the wiki vector space page. The conclusion one reaches is that the vector space properties are "induced" from V and W into Z. The fact that + is commutative within V and W causes + to be commutative within Z : z1 + z2 = v1w1 + v2w2 = (v1+v2) (w1+w2) = (v2+v1) (w2+w1) = v2w2 + v1w1 = z2 + z1 . Addition in Z is associative because it is associative in V and W: (z1+ z2) + z3 = ( v1w1 + v2w2) + v3w3 = (v1+v2)(w1+w2) + v3w3 = (v1+v2+v3) (w1+w2+w3) = v1w1 + (v2+v3)(w2+w3) = v1w1 + ( v2w2 + v3w3) = z1 + (z2 + z3) . The zero element in Z is 0 = 00 since vw + 0 = vw + 00 = (v+0)(w+0) = vw . The additive inverse of z = vw is -z = (-v)(-w) since z + (-z) = vw + (-v)(-w) = (v-v)(w-w) = 00 = 0 . For scalars a,b we have a(bz) = (ab)z compatibility since a(bz) = a(b[vw]) = a[ (bv)(bw) ] = (abv)(abw) = (ab)(vw) = (ab)z . Identity for scalar multiplication requires that 1(z) = z : 1(z) = 1(vw) = (1v)(1w) = vw = z . Distributive requirement #1: a(z1+z2) = az1+ az2 (a = scalar) 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 = az + bz (a,b = scalars) (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 QED 5. vw does not commute Fact: vw ≠ wv unless V = W and v = w. (B.12) Proof: V≠W: If V≠W, the object wv makes no sense since it would require w ϵ V and v ϵ W. V=W: vw - wv = vw + (-w)(-v) = (v-w)(w-v) ≠ 0 unless v = w. Compare (B.12) to the Fact stated in and below (4.1.1), Fact: vw ≠ wv unless V = W and v = w. (4.1.1) However: There is certainly an isomorphism between VW and WV. Writing VW ~ WV one could certainly then say that vw ~ wv . The same could be said for the operator. 6. Visualization of the Direct Sum Consider this example v = ϵ R2 w = ϵ R3 z = vw = = ϵ R5 . (B.13) Here we visualize the direct sum vector z as a tall column vector which is the stacking of the two smaller column vectors v and w. In the tall column vector, v and w each occupy a private region. Here then are the rules (B.3) and (B.2) : v1w1 + v2w2 = + = = (v1+v2) (w1+w2) (B.14) (αv1)(αv2) = = α = α(v1v2) . (B.15) The fact (B.10) that dim(VW) = dim(V) + dim(W) is demonstrated by 5 = 2+3. The fact (B.12) that vw ≠ wv is demonstrated since (a,b,s,t,u)T ≠ (s,t,u,a,b)T. 7. Extension to multiple products The axioms for a triple direct sum are these, v1w1x1 + v2w2x2 + ... + vkwkxk = (v1+v2 + ...+ vk) (w1+w2 + ... + wk) (x1+x2 + ... + xk) (B.16) (αv)(αw)(αx) = α(vwx) (B.17) and from this one can imagine an arbitrary number of involved in a direct sum. One can derive these two equations from (B.1) and (B.2) by assuming associativity and then grouping things for example as v1w1x1 + v2w2x2 + ... + vkwkxk = (v1w1)x1 + (v2w2)x2 + ... + (vkwk)xk = [(v1w1) + (v2w2) + ... + (vkwk)] (x1 + x2 + ... + xk) = [ (v1+v2+ .. +vk) (w1+w2+ .. +wk)] (x1 + x2 + ... + xk) = (v1+v2 + ...+ vk) (w1+w2 + ... + wk) (x1+x2 + ... + xk) and (αv)(αw)(αx) =[(αv)(αw)] (αx) = [α(vw)] (αx) = α [ (vw x] = α[vwx] . One can define Z ≡ VWX zi ≡ viwix1 ϵ Z (B.6)' : (V,W,X) → VWX : (v,w,x) ↦ (vwx) We leave it to the reader to prove the following extended claims: Lemma: Given some zi, we can find vi, wi and xi such that zi = viwixi . (B.8)' Fact: If dim(V) = n, dim(W) = n' and dim(X) = n". then dim(VWX) = n + n' + n". (B.10)' Fact: If V.W and X are vector spaces, then Z = VWX is a vector space. (B.11)' The extension of the "tall vector" visualization to the triple sum seems fairly obvious where one ends up stacking three vectors to make a single tall vector. 8. Application to tensor products Define the vector product space Vk as in (5.1). If T = V2V3 one can write t = Σij Tij eiej Σijk Tijk eiejek ϵ T // t = vw If T = V1V2V3 one can write t = ΣiTiei Σij Tij eiej Σijk Tijk eiejek ϵ T // t = vwx and in this manner we eventually arrive at (5.4.3) for T(V) T(V) ≡ V0 V V2 V3 ....... (5.4.2) t = s ΣiTi ei Σij Tij eiej Σijk Tijk eiejek ... ϵ T(V), s ϵ K (5.4.3) For the space V*k the objects being direct-summed are functionals instead of tensors, but the formalism is exactly the same, τ = s ΣiTi λi Σij Tij λiλj Σijk Tijk λiλjλk + ...... s ϵ K (6.4.3)