Hefferon J. A short linear algebra book (answers)(130s).pdf: eKnigu

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Hefferon J. A short linear algebra book (answers)(130s).pdf

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2 by the deﬁnition of ‘satisﬁes’. But, because k = 0, that’s true if and only if a1,1 s1 + a1,2 s2 + · · · + a1,n sn = d1 . . . and ai,1 s1 + ai,2 s2 + · · · + ai,n sn = di . . . and am,1 s1 + am,2 s2 + · · · + am,n sn = dm...

Answers to Exercises and then this computation works. u + v 2 + u − v 2 = (u1 + v1 )2 + · · · + (un + vn )2 + (u1 − v1 )2 + · · · + (un − vn )2 = u1 2 + 2u1 v1 + v1 2 + · · · + un 2 + 2un vn + vn 2 + u1 2 − 2u1 v1 + v1 2 + · · · + un 2 − 2un vn + vn 2 = 2(u1 2 + · · · + un 2 ) + 2(v1 2 + · · · + vn 2 ) =2 u 2+2 v 2...

where θ is the angle between the vectors. Thus the ratio is | cos θ|. 1.I I.2.29 So that the statement ‘vectors are orthogonal iﬀ their dot product is zero’ has no exceptions. 1.I I.2.30 The angle between (a) and (b) is found (for a, b = 0) with ab arccos( √ √ ). a2 b2 If a or b is zero then the angle is π /2 radians. Otherwise, if a and b are of opposite signs then the angle is π radians, else the angle is zero radians. 1.I I.2.31 The angle between u and v is acute if u v > 0, is right if u v = 0, and is obtuse if u v < 0. That’s because, in the formula for the angle, the denominator is never negative. 1.I I.2.33 Where u, v ∈ Rn , the vectors u + v and u − v are perpendicular if and only if 0 = (u + v) (u − v) = u u − v v, which shows that those two are perpendicular if and only if u u = v v. That holds if and only if u = v . 1.I I.2.34 Suppose u ∈ Rn is perpendicular to both v ∈ Rn and w ∈ Rn . Then, for any k , m ∈ R we have this. u (k v + mw) = k (u v) + m(u w) = k (0) + m(0) = 0 1.I I.2.35 We will show something more general: if z1 = z2 for z1 , z2 ∈ Rn , then z1 + z2 bisects the angle between z1 and z2 ¨ ¨ ¡ ¡ ¨¨ ¨¨ ¨ ¨ ¡ ¡ ¡ gives ! ¡ ¡ ¡ ¡ ¡ ¡ B ¨ ¨ ¡ ¨ ¡ ¨ ¡¨ ¡¨ ¨ ¨...

and any cosmetic change, like multiplying the bottom row by 2,   21 1 3 0 1 −2 −7 0 0 19 70 gives another. 1.I I I.1.13 (a) The ρi ↔ ρi operation does not change A. (b) For instance, l 1 − ρ1 +ρ1 2 −→ 34 eaves the matrix changed. (c) If i = j then   . .  . ai,1 · · · ai,n     .  . .   aj,1 · · · aj,n    . . . ...

Linear Algebra, by Hefferon r (a) Every such set has the form {r · v + s · w , s ∈ R} where either or both of v, w may be 0. With the inherited operations, closure of addition (r1 v + s1 w) + (r2 v + s2 w) = (r1 + r2 )v + (s1 + s2 )w and scalar multiplication c(rv + sw) = (cr)v + (cs)w are easy. The other conditions are also routine. (b) No such set can be a vector space under the inherited operations because it does not have a zero element. 2.I.1.39 Yes. A theorem of ﬁrst semester calculus says that a sum of diﬀerentiable functions is diﬀerentiable and that (f + g ) = f + g , and that a multiple of a diﬀerentiable function is diﬀerentiable and that (r · f ) = r f . 2.I.1.40 The check is routine. Note that ‘1’ is 1 + 0i and the zero elements are these. 2 + 2(a) (00 0i) + (0 + 0i)x + (0 + 0i)x + 0i 0 + 0i (b) 0 + 0i 0 + 0i .I.1.41 Notably absent from the deﬁnition of a vector space is a distance measure. 2.I.1.43 (a) We outline the check of the conditions from Deﬁnition 1.1. Item (1) has ﬁve conditions. First, additive closure holds because if a0 + a1 + a2 = 0 and b0 + b1 + b2 = 0 then (a0 + a1 x + a2 x2 ) + (b0 + b1 x + b2 x2 ) = (a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 is in the set since (a0 + b0 ) + (a1 + b1 ) + (a2 + b2 ) = (a0 + a1 + a2 ) + (b0 + b1 + b2 ) is zero. The second through ﬁfth conditions are easy. Item (2) also has ﬁve conditions. First, closure under scalar multiplication holds because if a0 +a1 +a2 = 0 then r · (a0 + a1 x + a2 x2 ) = (ra0 ) + (ra1 )x + (ra2 )x2 is in the set as ra0 + ra1 + ra2 = r(a0 + a1 + a2 ) is zero. The second through ﬁfth conditions here are also easy. (b) This is similar to the prior answer. (c) Call the vector space V . We have two implications: left to right, if S is a subspace then it is closed under linear combinations of pairs of vectors and, right to left, if a nonempty subset is closed under linear combinations of pairs of vectors then it is a subspace. The left to right implication is easy; we here sketch the other one by assuming S is nonempty and closed, and checking the conditions of Deﬁnition 1.1. Item (1) has ﬁve conditions. First, to show closure under addition, if s1 , s2 ∈ S then s1 + s2 ∈ S as s1 + s2 = 1 · s1 + 1 · s2 . Second, for any s1 , s2 ∈ S , because addition is inherited from V , the sum s1 + s2 in S equals the sum s1 + s2 in V and that equals the sum s2 + s1 in V and that in turn equals the sum s2 + s1 in S . The argument for the third condition is similar to that for the second. For the fourth, suppose that s is in the nonempty set S and note that 0 · s = 0 ∈ S ; showing that the 0 of V acts under the inherited operations as the additive identity of S is easy. The ﬁfth condition is satisﬁed because for any s ∈ S closure under linear combinations shows that the vector 0 · 0 + (−1) · s is in S ; showing that it is the additive inverse of s under the inherited operations is routine. The proofs for item (2) are similar....

24 and note that the resulting linear system a1,1 c1 + a1,2 c2 + a1,3 c3 + a1,4 c4 + a1,5 c5 = 0 a2,1 c1 + a2,2 c2 + a2,3 c3 + a2,4 c4 + a2,5 c5 = 0 a3,1 c1 + a3,2 c2 + a3,3 c3 + a3,4 c4 + a3,5 c5 = 0 a4,1 c1 + a4,2 c2 + a4,3 c3 + a4,4 c4 + a4,5 c5 = 0...

shows that the original system is nonsingular if and only if the 3, 3 entry is nonzero. This fraction is deﬁned because of the ae − bd = 0 assumption, and it will equal zero if and only if its numerator equals zero. We next worry about the assumptions. First, if a = 0 but ae − bd = 0 then we swap     1 b/a c/a 0 0 1 b/a c/a ρ2 ↔ρ3 0 0 (af − cd)/a 0 −→ 0 (ah − bg )/a (ai − cg )/a 0 0 (ah − bg )/a (ai − cg )/a 0 0 0 (af − cd)/a 0...

Answers to Exercises (a) Setting gives this (L2 M 1 T −2 )p1 (L0 M 0 T −1 )p2 (L3 M 0 T 0 )p3 = (L0 M 0 T 0 )...

45 ∼ M3×2 . (c) No, for instance, M2×3 = 3.I.2.22 One direction is easy: if the two are isomorphic via f then for any basis B ⊆ V , the set D = f (B ) is also a basis (this is shown in Lemma 2.3). The check that corresponding vectors have the same coordinates: f (c1 β1 + · · · + cn βn ) = c1 f (β1 ) + · · · + cn f (βn ) = c1 δ1 + · · · + cn δn is routine. For the other half, assume that there are bases such that corresponding vectors have the same coordinates with respect to those bases. Because f is a correspondence, to show that it is an isomorphism, we need only show that it preserves structure. Because RepB (v ) = RepD (f (v )), the map f preserves structure if and only if representations preserve addition: RepB (v1 + v2 ) = RepB (v1 ) + RepB (v2 ) and scalar multiplication: RepB (r · v ) = r · RepB (v ) The addition calculation is this: (c1 + d1 )β1 + · · · + (cn + dn )βn = c1 β1 + · · · + cn βn + d1 β1 + · · · + dn βn , and the scalar multiplication calculation is similar. 3.I.2.23 (a) Pulling the deﬁnition back from R4 to P3 gives that a0 + a1 x + a2 x2 + a3 x3 is orthogonal to b0 + b1 x + b2 x2 + b3 x3 if and only if a0 b0 + a1 b1 + a2 b2 + a3 b3 = 0. (b) A natural deﬁnition is this.   a1 a0 2a2  a1  D( ) =   3a3  a2  a3 0 3.I.2.25 Because V1 ∩ V2 = {0V } and f is one-to-one we have that f (V1 ) ∩ f (V2 ) = {0U }. To ﬁnish, count the dimensions: dim(U ) = dim(V ) = dim(V1 ) + dim(V2 ) = dim(f (V1 )) + dim(f (V2 )), as required. 3.I.2.26 Rational numbers have many representations, e.g., 1/2 = 3/6, and the numerators can vary among representations....