Solutions 2

# Solutions 2 - MAS212 Linear Algebra I Exercises 2 (with...

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Unformatted text preview: MAS212 Linear Algebra I Exercises 2 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday 13th 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. [30] In the following, prove that U either is or is not a vector subspace of the speciﬁed vector space. (a) Let U ={(x, y, z) E R3 | x + y = 0}. Is U a vector subspace of R3? (b) Let U={(x, y, z) E [R3 | x +y =1}. Is U a vector subspace ofR3? (c) Let U={(w, x, y, z) E 1R4| w +x =0,y +z=0}. Is Ua vector subspace ofR“? ((1) Let U={(x, y) E R2 | x =iy }. Is Ua vector subspace oflRZ? (e) Let U={(x, y) E R2 I x2—y2 =1}. ls Ua vector subspace oflRZ? (i) Let U={(x,y) E R2 | x =y = 0}. Is U a vector subspace ofﬂz? Solution It is necessary to check the vector subspace conditions, in one form or another. (a) Yes, because the constraint is homogeneous linear. (0, 0, 0) satisﬁes the constraint since 0 + 0 = 0, so U contains the zero vector. To check closure, let u = (x1,y1,zl), v= (x2, y2, 22) E U. Thenx1 +y1 =0 (l),x2 +y2=0 (2). The vector u + v = (x1 + x2, y1 + y2, z1 + 22) E U if it satisﬁes the constraint on U that (x1 +x2) + (y1 +y2) =0. Adding constraints (1) and (2) gives (x1 +x2) + (y1 +y2) =0. Hence u + v E U and U is closed under addition. For any a E R, the vector a u = (a x1,ay1,a 21) E Uifit satisﬁes the constraint on Uthat (a x1) + (ayl) = 0. Multiplying constraint (1) by a gives (a x1) + (a y1) = 0. Hence a u E U and U is closed under scalar multiplication. Therefore, U is a vector subspace of R3. (b) No, because the constraint is not homogeneous linear. In particular, (0, 0, 0) E U since 0 + 0 ab 1, so U does not contain the zero vector. (0) Yes, because the constraints are homogeneous linear. (0, 0, 0, 0) satisﬁes the constraints because 0 + 0 = 0, 0 + 0 = 0, so U contains the zero vector. To check closure, let u = (W1, x1,y1, 21), v= (w2,x2, y2, 22) E U. Then w1+x1=0,y1+zl=0 (1), W2 +x2=0,y2 +zz=0 (2). The vector u + v = (w1 + wz, x1 + x2, y1 + yz, z1 + 22) E U if it satisﬁes the constraint on U that (w1+ wz) + (x1+ x2) =0, (y1+y2) + (21+22) =0. Adding constraints (1) and (2) gives (w1+w2) + (x1+x2) =0, (y1+y2) + (21+22) =0. Henceu +v E Uand Uis closed under addition. For any a E R, the vector a u = (a w1 , a x1, a yl, a 21) E U if it satisﬁes the constraint on Uthat (a wl) + (a x1) = 0, (ayl) + (a 21) = 0. Multiplying constraint (1) by a gives (a wl) +(ax1)=0,(ay1) +(azl)=0. Hencea u E Uand Uis closed under scalar multiplication. Therefore, U is a vector subspace of R4. (d) No, because x =i y is a nonlinear constraint equivalent to x2 = y2. The solution set of this equation is cross-shaped. Alternatively, it means + or —, so U={(x,y) E R2|x =yorx = —y}={(x,y) E R2 | x =y}U {(x,y) E R2 | x = —y}andingenerala union of vector subspaces is not a vector subspace. l (0, 0) satisﬁes the constraint because 0 =i 0, so U contains the zero vector. But a counterexample shows that U is not closed under addition: e.g. (1, 1), (1,—1) E U but (1,1) +(1,—l)=(2,0) GE Usince2 abiO. (e) No, because the constraint is nonlinear. In particular, (0, 0) \$ U since 0—0 at 1, so U does not contain the zero vector. (1) Yes, because the constraints, which could also be written as x = 0, y = 0, are homogeneous linear. In fact, U = {(0, 0) E IRZ}, which is the zero vector space as a vector subspace of R2. A more formal proof that U is a vector subspace could easily be written as in the examples above. 2. [20] Let Vbe a vector space over a ﬁeld [K and let u and v be ﬁxed elements of V. Consider the set U = {a u + b v| a, b E K}. Show that (a) U is a subset of Vand (b) U is a vector subspace of V. List the elements of U when K= F2. Solution (a) By the vector space closure axioms, a u + b v E V. Hence U E V. (b) Taking a = 0, b = 0 implies 0V 6 U so U contains the zero vector. If a1, a2, b1, b2 E K then alu +b1v,a2u +b2v E Vand(a1u +b1v)+(a2u +b2v)= (a1 +a2) u + (b1+b2)v E U since it has the required form because (a1 + a2) , (b1 + b2) 6 K so U is closed under vector addition. Ifk E Kthen k(a u + b v) = (ka) u + (kb) v E Usince it has the required form becauseka, k b E K, so U is closed under scalar multiplication. Therefore, U is a vector subspace of V. When K= 0:2 = {0, 1}, enumerating all possible values of a and b shows that U = {0, u, v, u + v}. 3. [20] Let V=MZ’ 2( R) be the vector space of real 2 x 2 matrices with the usual addition and multiplication by a scalar. Let ab d0 Cd U1= a,b,cElR,U2= delR. Prove that U1 and U2 are vector subspaces of V. What would you guess their dimensions to be and why? [We will see how to determine dimension formally in Chapter 4.] Solution U1: Choosing a = b = c = 0 shows that the zero matrix is an element of U1. Consider two elements of U1 with matrix elements denoted by subscripts 1 and 2. Then a1 b1 a2 b2 czl+a2 b1+b2 + = c1 a1 c2 a2 c1+c2 al+a2 and for allk E [R a b ka kb kl = c a kc ka Both right hand sides have the required form to be elements of U1, so U1 is closed under vector space operations. Hence U1 is a vector subspaces of V. The general element of U1 contains 3 parameters, so guess that U1 has dimension 3. U2: Choosing d = 0 shows that the zero matrix is an element of U2. Consider two elements of U2 with matrix elements denoted by subscripts 1 and 2. Then d1 0 d2 0 d1+d2 o 0111+ 0—d2= 0 —d1—d2 and for allk E R d 0 kd o ' o —d = o —kd Both right hand sides have the required form to be elements of U2, so U2 is closed under vector space operations. Hence U2 is a vector subspaces of V. The general element of U2 contains 1 parameter, so guess that U2 has dimension 1. 4. [30] LetS = {(x, y, z) E R3 | x = 2y}, T= {(x, y, z) E R3 | y = -z} be vector subspaces of the vector space R3. (a) Show that S n T = {(x, y, z) E R3 | x = 2y, y = —z} and hence prove that S n T is a vector subspace of the vector space R3. (b) Find a counterexample that proves that SU T is not a vector subspace of the vector space R3. (c) Show that S and Tcan also be written as S ={(2y, y, z) E R3 | y, z E R}, T={(x, -z', z') E R3 | x, 2'6 R}, where z' is used in Tto show that is is not related to thez used in S. Hence, express S + T in a similar form, namely {(a, b, c) E R3 | y, z, x, 2'6 R}, where you must ﬁnd expressions fora, b, c. By ﬁnding values of y, Z. x, 2' such that (a, b, c) = (X, Y, Z) for an arbitrary vector (X, Y, Z) E R3, prove that S + T = R3 and is therefore a vector subspace of the vector space R3. Solution (a) The intersection of S and Tis the set of all elements that satisfy both the constraint on S and the constraint on T, so the result for S0 T follows; the comma separating multiple constraints can be interpreted as a logical "and". Let U = {(x, y, z) E R3 | x = 2y, y =—z}. Prove that U is a vector subspace of R3. (0, 0, 0) E Usince 0 =2-0, 0 =—0. If (x1,y1, 21), (x2,y2, 22) E Uthenx1=2y1,y1=—z1 and x2 = 2y2, y2 = —22. Adding the constraints shows that (x1+x2) =2(y1+y2), (y1+y2) =—(z1 +22), so (x1+x2,y1+y2, 21+22) E Usince it satisﬁes the constraints on U. (b) (2,1, 0) 6 Sand (0,—1, 1) E T, so (2, 1,0), (0,—1, 1) 6 SU T,but (2, 1, 0) + (0,—1, l) = (2, 0,1) 65 SU T since (2, 0,1) 65 Sand (2, 0,1) \$ T. Therefore SU Tis not closed under addition and so is not a vector subspace. (c) Substituting the constraints into the general vectors to eliminate one of the parameters gives the forms shown. Note that the parameters can be renamed without changing the deﬁnitions of the subsets. Then the deﬁnition of the sum of two vector subspaces leads immediately to S + T={(2y +x,y—z§ z+z’) E R3 |y,z, x, 2'6 R}. Now set (2y +x,y—z§ z+z') = (X, Y, Z) and solve fory, z, x in terms ofX, Y, Z, 2' to givex =X—2( Y+z'),y = Y+z£ z = Z—z'. This means that for any value of z’we can ﬁnd values for y, z, x and hence every vector (X, Y, Z) 6 R3 is also in S + T; therefore S + T = R3. (More precisely, S E R3 and T E R3 by deﬁnition so S + T E R3, and we have just shown that every vector in R3 is also in S + T so R3 E S + T therefore S + T = R3.) This analysis looks forward to the deﬁnition of spanning sets next week. ...
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## This note was uploaded on 02/11/2012 for the course MTH 141, 142, taught by Professor Mcallister during the Spring '08 term at SUNY Empire State.

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Solutions 2 - MAS212 Linear Algebra I Exercises 2 (with...

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