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Solutions 3 - MAS212 Linear Algebra I Exercises 3(with...

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Unformatted text preview: MAS212 Linear Algebra I Exercises 3 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday 20th October 2006 in the blue box on the ground floor. Your submission must have your FULL NAME and STUDENT NUMBER PRINTED CLEARLY at the top and separate sheets must be stapled together. Marks out of 100 are shown in square brackets after each question number. 1. [20] In the vector space [R3, determine whether each of the vectors u = ( 1, 2, 0) and v = (0, 1, 2) is in the vector subspace spanned by (1, 2, 3) and (2, l, 0). V Solution For u, try to solve (1, 2, 0) = a (l, 2, 3) + b (2, l, 0) for a, b E R Split the vector equation into three coupled scalar equations to give 1=a+2b, 2=2a+b, 0=3a. The last equation implies that a = 0. Then the first two equations become 1 = 2 b, 2 = b, which are clearly inconsistent (since 1 at 4), so there are no solutions. Therefore, u 65 ((1, 2, 3), (2, l, 0)). For v, try to solve (0, 1, 2) = a (1, 2, 3) + b (2, l, 0) for a, b 6 IR. Splitting the vector equation into three coupled scalar equations gives 0 = a + 2 b, l = 2 a + b, 2 = 3 a. The last equation gives a = A. Then the first equation gives b = —£ = —i. This solution is 3 2 3 consistent with the second equation since 1 = 2(1 j + {—i J. Therefore, 3 3 L v 6((1,2,3), (2,1,0)>- > Maple Solution 2. [20] (a) Let v1 = (1, 2), v2 = (2, l) E R2. Show that {V1, v2} is a linearly independent set of vectors. (b) Letv1= (1, O, 2), v2= (2, 1, —l ), v3 =(1,1,1) E R3. Show that {V1, v2, v3} is a linearly independent set of vectors. (c) Let v1 = (l, 0, —l, —l ), v2 = (0, 0, 1,—1), v3 = (0,—1, 0, 0), v4= (—2, 2,1, 3) E R4. Show that {V1, v2, v3, v4} is a linearly dependent set of vectors. V Solution (a) To determine whether the vectors are linearly independent we need to solve a (1, 2) + b (2, 1) = 0 for a, b E R This vector equation is equivalent to the two scalar equations a + 2 b = 0, (1) 2 a + b = 0. (2) (1) => a = —2 b. Substituting for a in (2) gives -4 b + b =—3 b = 0, hence the only solution is a = b = 0. Therefore the vectors are linearly independent. (b) Solvea(1,0,2) +b(2,1,—1)+c(1,1,1)=0fora,b,cER,i.e. a+2b+c=0, (l) b+c=0, (2) 2a—b+c=0. (3) 1 (2) = b =—c. Substituting for b in (1) and (3) gives a—2 c + c = 61-6 = 0 = a = c and 2 a + c + c=2 a + 2 c=0 => a =—c. Hence the only solution is a =b =c=0. Therefore the vectors are linearly independent. (c) Solvea (l, 0, —1, —1)+b(0, 0,1,—1)+c(0,—1, 0, 0) +d (—2, 2,1, 3) =0 fora, b, c, d E IR, 1.e. a —2 d = 0, (l) —c + 2 d = 0, (2) —a + b + d = 0, (3) —a—b +3 d=0. (4) (1 ) = a =2 dand (2) => c =2 d. Substituting for a in (3) and (4) givesb —d=0, —b + d=0. These two equations are equivalent and imply that b = d, so the solution is a = 2 d, b = d, c = 2 d for any d E R Clearly a = b = c = d = 0 is a solution, but this solution is not unique and there are i nonzero solutions for any nonzero d E R Therefore the vectors are linearly dependent. > Maple Solution 3. [20] In the vector space [R3, (a) find a maximal (i.e. as large as possible) linearly independent subset ofthe set ofvectors {(1,—2, 1), (2, 1, 2), (1, —1, 1), (1, 1, 1)}, and (b) extend it to a linearly independent set of three vectors and show that this forms a basis for R3. V Solution (a) There cannot be more than 3 linearly independent vectors in R3, so we can discard at least one. Byinspection,(1,—1,1)+(1,1,1)=(2, 0, 2) and (1, 1, 1) —(l,—1, 1) = (0, 2, 0). From these two vectors we can construct the first two vectors in the set as (1,—2, 1) = i(2, 0,2) —(0, 2, 0) and (2, 1, 2) = (2, 0, 2) + %(0, 2, 0). Moreover, 2 { (1, —l, l ), (l, l, l )} is clearly a linearly independent set since one vector is not a scalar multiple of the other, and since the other two vectors are linear combinations of these two vectors, this is a maximal linearly independent set. (b) Try including the first standard basis vector for R3 to give {(1, —1, 1), (l, l, l), (l, 0, 0)}. To prove that this set is linearly independent solvea( l, —l, l) + b( l, l, l) + c( l, 0, 0) = 0 for a, b, c E R, Le. a +b + c=0, (l) —a +b=0, (2) a +b=0. (3) (2) + (3) => 2b=0 => b=0 and (3)—(2) => 2a=0 => a=0. Then(1) => c=0. Hence a = b = c = 0 uniquely and so the vectors are linearly independent. To determine whether they also spaan3we must solvea(l,—1, 1) +b(1, 1, 1) +c(1, 0, 0) = (x,y,z) E R3 fora, b, c E IR, i. e. a +b + c=x, (l) -a+b=y, (2) a+b=z. (3) (2) +(3) =>2b=y+z=>b=‘m and(3)—(2) =>2a=z—y=>a= z—y- Then 2 2 (1) = c=x—(o + b) =x—z. Hence there are solutions fora, b, c E Rfor anyx,y, z E Rand so , the set ofvectors spans R3. Therefore { (l, —1, 1), (l, l, l), (l, 0, 0)} forms a basis for #33. > Maple Solution 4. [20] (a) Let v1 = (1, 1) and v2 = (l, 2). Show from first principles that {v1, v2} forms a basis for 032. (b) Letv1= (1, 2), v2= (2, l) and v3 = (l, 1). Show that {v1, v2, v3} is a spanning set for #32. Is it also a basis? (e) Let v1 = (0, 1, 0), v2= (1, 2,1), v3 = (0,—1, 1) and v4= (l, 0, —1 ). Show that {v1, v2, v3, v4} is a spanning set for R3. Is it also a basis? V Solution (a) A basis is a linearly independent spanning set. Suppose a (1, 1) + b (1, 2) = 0. This corresponds to the two scalar equations a + b = 0, a + 2 b = 0 => a = 0, b = 0. Therefore v1, v2 are linearly independent. Suppose (x, y) = a (1, 1) + b (1, 2). This corresponds to the two scalar equationsx =a + b, y = a + 2 b = a = 2 x—y, b =y—x. Therefore any (x, y) 6 IR2 can be expressed as (x,y) = (2 x—y) (1,1) + (y—x) (1, 2), so {V1, v2} spans R2 and therefore forms a basis for R2. (b) Suppose (x,y) =a (1,2) +b (2, l) +c (1, 1) => x=a +2b +0, (1) y=2a+b+c. (2) We have only 2 equations, so we can solve for only 2 unknowns, say a, b, so regard c as a parameter. 2 _ _ 2 - (1) — (2) = 2x—y=3 b +c = 3 b=2x—y—c = b = %. Substitutingthis into _ 2 _ (1) =>a =x—2 b—c =x-% (2 x—y—c) -c = $ This solution holds v x, y, c 6 IR, therefore {v1, v2, v3} is a spanning set for R2. However, it is not a basis because it cannot be linearly independent. To prove this, setx = y = 0 in the above equations to give a = —Tc , b = _Tc V c E IR, which is not unique and gives nonzero solutions for any c at 0. (c) Suppose(x,y,z)=a (0,1,0) +b (1,2,1) +c (0,—1,1) +d(1,0,—1) => x = b + d, (l) y=a +2 b—c, (2) z = b + c—d. (3) Solving for a, b, 0 regarding d as a parameter, (1) = b =x—d. Then (3) => c=z—b +d=z—x +2 dand (2) => a =y—2 b + c=y—2x + 2 d +z—x +2 d =y—3 x +4 d +z. This solution holds V x, y, z, d E IR, therefore {V1, v2, v3, v4} is a spanning set for IR5. However, it is not a basis because it cannot be linearly independent. To prove this, setx = y =z = 0 in the above equations to give a = 4 d, b = —d, c = 2 d V d E IR, which is not unique and gives nonzero solutions for any , d at 0. > Maple Solution 5. [20] Find a basis for U, where: (a) U={(x, y, z) | x, y, z E R, x + 2y—z= 0} is a vector subspace ofIR3; (b) U={(W, x, y, Z) I W, x, y, Z G R, x—y= 0, w +z=0} is a vector subspace ofIR4. V Solution (a) Solving the constraint forx (say) gives x = —2 y + 2. Therefore, U={(—2y +z,y,z) | y,z E IR}={y (—2, 1, 0) +2 (1, 0,1)| y,z E R}, so{(—2, 1, 0), (1, 0,1)}is a spanning set for U. It can easily be proved to be a linearly independent set (since one vector is not a scalar multiple of the other), therefore it is a basis for U. (b) Solving the constraints forx, w (say) gives x = y, w = —z. Therefore, U={(-z,y,y,z)| y,z E R}={y(0,1,1,0)+z(—l,0,0,l)|y,z 6 IR}, so {(0, l, l, 0), (—l, 0, 0, 1)} is a spanning set for U. It can easily be proved to be a linearly 3 L independent set (since one vector is not a scalar multiple of the other), therefore it is a basis for U. ...
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