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Math 315: Linear Algebra Solutions to Assignment 5 #1 Determine whether a given set of objects together with operations form a vector space. (a) the set of triples (x, y, z) with (x, y, z) + (x0 , y 0 , z 0 ) = (x + x0 , y + y 0 , z + z 0 ), " (b) (c) (d) and k(x, y, z) = (kx, y, z); # a 1 with matrix addition and scalar multiplication; 1 b " # a 0 the set of all 2×2 matrices of the form with matrix addition and scalar multiplication; 0 b the set of all positive real numbers with the operations the set of all 2×2 matrices of the form x + y = xy, and kx = xk . Solution (a) is not a vector space, since (k + l)u 6 ku + lu (Axiom 8). (b) is not a vector space, because the sum of two such matrices does not result in a matrix with 1’s on the diagonal (Axiom 1). (c) and (d) are vector spaces, since all axioms are satisfied (this requires some work to show, though). #2 Prove: For any vectors u, v, and w, the vectors u − v, v − w, and w − u form a linearly dependent set. Solution The sum (which is a non-trivial linear combination) of those vectors results in the zero vector: (u − v) + (v − w) + (w − u) = 0 so the vectors in question are linearly dependent. 1 #3 Show that every set with three or more vectors from R2 is linearly dependent. Solution Let the vectors be v1 , v2 , and v3 . If v1 = kv2 for some k 6= 10, then 1 · v1 − k · v2 + 0 · v3 = 0, so the vectors are linearly dependent. Assume v1 and v2 are not multiples of each other. Then, as we did several times in class, any other vector of R2 can be represented as their linear combination. In particular, v3 = k1 v1 + k2 v2 for some k1 , k2, so (−k1 ) · v1 + (−k2 ) · v2 + 1 · v3 = 0, and the vectors are linearly dependent too. #4 Which of the following are linear combinations of " A= " (a) 6 −8 −1 −8 # 4 0 −2 −2 " (b) 0 0 0 0 # " B= # " (c) 1 −1 2 3 6 0 3 8 # " C= # 0 2 ? 1 4 # Solution All of them are linear combinations: " 6 −8 −1 −8 " " # = 1·A+2·B−3·C 0 0 0 0 # 6 0 3 8 # = 0·A+0·B−0·C = 1·A+2·B+1·C #5 Use Theorem 5.2.1 to determine which of the following are subspaces of the space F (−∞, ∞). (a) all f such that f (0) = 0 (b) all f of the form k1 + k2 sin x, where k1 and k2 are real numbers. 2 Solution Let V be the set of objects in question. We have to check the conditions (a) and (b) of the Theorem. (a) For f, g ∈ V one has (f + g)(0) = f (0) + g(0) = 0 + 0 = 0, so the first condition is satisfied. Also, (kf )(0) = 0, so the second condition is satisfied. (b) Let f (x) = k10 + k20 sin x and g(x) = k100 + k200 sin x for some numbers k10 , k20 , k100 , k200 . One has (f + g)(x) = (k10 + k100 ) + (k20 + k200 ) sin x, so the sum of functions is in V . Similarly, (kf )(x) = kk10 + kk20 sin x ∈ V , so V is a vector space. #6 Use Theorem 5.2.1 to determine which of the following are subspaces of the space Mnn . (a) all n × n matrices A such that AT = −A (b) all n × n matrices A such that AB = BA for a fixed n × n matrix B. Solution Again, we have to check the conditions (a) and (b) of the Theorem. (a) Let A0 , A00 ∈ V . This implies (A0 )T = −A0 and (A00 )T = −A00 . Then (A0 + A00 )T = (A0 )T + (A00 )T = −A0 − A00 = −(A0 + A00 ), so the first condition is satisfied. Similarly, (kA0 )T = k(A0 )T = k(−A0 ) = −(kA0 ), so the second condition is satisfied as well. (b) Let A0 , A00 ∈ V . This implies A0 B = BA0 and A00 B = BA00 . One has (A0 + A00 )B = A0 B + A00 B = BA0 + BA00 = B(A0 + A00 ), so the first condition is satisfied. Similarly, (kA0 )B = k(A0 B) = k(BA0 ) = B(kA0 ), so the second condition is satisfied as well. #7 Express 7+8x+9x2 as a linear combination of p1 = 2+x+4x2 , p2 = 1−x+3x2 , and p3 = 3+2x+5x2 . Solution 7 + 8x + 9x2 = 0 · p1 − 2 · p2 + 3 · p3 . 3 #8 Let f = cos2 x and g = sin2 x. Which of the following lie in the space spanned by f and g? (a) cos 2x (b) 3 + x2 (c) 1 (d) sin x (e) 0 Solution cos 2x = cos2 x − sin2 x = f − g 1 = cos2 x + sin2 x = f + g 0 = 0 · cos2 x + 0 · sin2 x = 0 · f + 0 · g so cos 2x, 1, and 0 are in span(f , g). To show that 3 + x2 6∈ span(f , g), assume the contrary, that is 3 + x2 = k1 · cos2 x + k2 · sin2 x for some numbers k1 , k2 . However, there is no way to satisfy the above equality for all x: if x → ∞, then 3 + x2 → ∞, and the sum k1 · cos2 x + k2 · sin2 x is bounded by |k1 | + |k2 |. To show that sin x 6∈ span(f , g), assume the contrary, that is sin x = k1 · cos2 x + k2 · sin2 x for some numbers k1 , k2 and all x. But then sin(−x) = k1 · cos2 (−x) + k2 · sin2 (−x) = k1 · cos2 (x) + k2 · sin2 (x) = sin x, which is a contradiction. #9 Use Theorem 5.2.4 to show that v1 = (1, 6, 4), v2 = (2, 4, −1), v3 = (−1, 2, 5) and w1 = (1, −2, −5), w2 = (0, 8, 9) span the same subspace of R3 . Solution To show this we have to check that w1 and w2 are linear combinations of v1 , v2 , v3 : w1 = 0 · v1 + 0 · v2 − 1 · v3 w2 = 2 · v1 − 1 · v2 + 0 · v3 so both groups of vectors span the same space. 4 #10 Find an equation for the plane spanned by the vectors u = (−1, 1, 1) and v = (3, 4, 4). Solution Since span(u, v) is a linear subspace of R3 , it should contain the zero vector, which means that in the plane equation ax + by + cz + d = 0 the coefficient d is 0. Therefore, the plain is of the form ax + by + cz = 0. We know that each of the given vectors is in the plain, so it satisfies the plain equation. This leads to the system of linear equations for determining a, b, c: −a + b + c = 0 3a + 4b + 4c = 0 which has a solution a = 0, b = −t, and c = t. Choosing c = 1 results in the plain equation −y + z = 0. 5