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3 Show that if A is an m×n matrix, then the solution set V to the equation Ax = 0 is a subspace of Rn Solution A1) Let x1;x2 ∈ Rn be two solutions to the equation Ax = 0 (that is, x1;x2 ∈ V)Then x1 x2 ∈ Rn, and A(x1 x2) = Ax1 Ax2 = 00 = 0 Thus x1 x2 ∈ V M1) Let x1 ∈ V, k ∈ R Then kx1 ∈ Rn, and A(kx1) = k(Ax1) = k(0) = 0 Thus kx1 ∈ V as well Thus by the subspaceA vector space V is a collection of objects with a (vector) addition and scalar multiplication defined that closed under both operations and which in addition satisfies the following axioms (i) (αβ)x = αxβx for all x ∈V and α,β∈F (ii) α(βx)=(αβ)x (iii) xy = y x for all x,y ∈V (iv) x(y z)=(xy)z for all x, y, z ∈VD u v Z l v } v o Z l v µ v d Z u u v µ Ç Á Z À } } v Z Á v µ X í ó X ì ì U ^ Z u î ì X ì ì



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Proposition 1 If X,Y ∈ Rn are orthogonal sets then either they are disjoint or X ∩Y = {0} Proof v ∈ X ∩Y =⇒ v ⊥ v =⇒ v·v = 0 =⇒ v = 0 Proposition 2 Let V be a subspace of Rn and S be a spanning set for VA!) 5 All the ma rgina ls (dime nsio n les s tha n p ) o f X ar e (m ultiv aria te) nor mal, but it is p ossible in theo ry to ha v e a collectio n o f un iva riate nor mals w ho se join t distributio nℓ Rn → R, ℓ(v) = v·v0, where v0 ∈ Rn ℓ(xy) = (xy)·v0 = x·v0 y·v0 = ℓ(x)ℓ(y), ℓ(rx) = (rx)·v0 = r(x·v0) = rℓ(x) • Cross product with a fixed vector L R3 → R3, L(v) = v×v0, where v0 ∈ R3 • Multiplication by a fixed matrix L Rn → Rm, L(v) = Av, where A is an m×n



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