Solution to Set 6, Math W = {x : x 1 + x 2 = 0} Solution: This is not a subspace since it does not contain 0 = (0, 0) since

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1 Solution to Set 6, Math , No.. Determine whether W is a subspace of R and give, where W = {x : x x = } Solution: This is not a subspace since it does not contain = (, ) since. 3., No. 6. Determine whether W is a subspace of R and give a geometric description of W, where W = {x : x + x = } Solution: The only point in R is = (, ) since condition x + x = implies x =, x =. Since + = W and for any scalar c, c = W, it follows that both closure properties C, C hold and W is a subspace. 3., No. 3. Determine whether W is a subspace of R 3. If W is a subspace, then give geometric description of W. W = { x : x = x + x } Solution: Note that (, ) W since = + and (, ) W since = + where as (, ) + (, ) = (3, ) / W since Hence property C is not satisfied and W is not a subspace. 3., No. 8. Let a is a given vector in R 3 and define W to be the subset R 3 such that W = { x : a T x = } Prove that W is a subspace of R 3. Solution: Assume u, v W, then we know a T u = = a T v. Therefore, we note that a T (u + v) = a T u + a T v =. Hence u + v W (property C checked). Now, if u W and c is an arbitrary constant, then we note that a T (cu) = ca T u = since a T u = as u W. Thus property C holds. Since both properties C, C hold, we have W to be a subspace of R 3. 3., No.. Let W be the subset f R 3 so that for fixed vectors a, b R 3, we have W = { x : a T x =, b T x = } Solution: Assume u, v W. Then we know a T u = = a T v, b T u = = b T v. Therefore, we note that a T (u + v) = a T u + a T v =. and b T (u + v) = b T u+b T v =. Hence u+v W (property C checked). Now, if u W and c is an arbitrary constant, then we note that a T (cu) = ca T u = since a T u = ; also b T (cu) = cb T u = since b T u = ; also as u W. Hence cu W and property C is checked. Since both properties C, C hold, we have W to be a subspace of R 3. 3., No. 6. In R 4 let x = [, 3,, ] T, y = [,, 3, ] T and z = [ 3,,, 4] T. Let a = ad b = 3. Illustrate that all ten properties of Theorem are satisfied by x, y, z, a, b.

2 Property C Property C Property a 3 x + y = = 5 R4 3 ax = 3 = y + x = = 5 R4 = x + y 3 Property a Since for ordinary numbers we know that x + (y + z ) = (x + y ) + z we have 3 x + (y + z) = = 4 = (x + y) + z 4 7 Property a3 Clearly = [,,, ] T R n has the property that for any x including the x provided x + = 3 + = 3 = x Property a4 We note that x = [, 3,, ] T = [, 3,, ] T has the property that x + ( x) = [, 3,, ] T + [, 3,, ] T = [,,, ] T =. Property m a(bx) = ( 3x) = [ 3, 9, 6, 3] T = [ 6, 8,, 6] T. On the otherhand (ab)x = ( 6)[, 3,, ] T = [ 6, 8,, 6] T = a(bx). Property m a(x + y) = ( [, 3,, ] T + [,, 3, ] ) T = [3,, 5, 3] T = [6,,, 6] T where as ax+ay = [, 3,, ] T +[,, 3, ] T = [, 6, 4, ] T +[4, 4, 6, 4] T = [6,,, 6] T = a(x+y) So property m holds. Property m3 (a + b)x = ( 3)x = x = [, 3,, ] T while ax + bx = [, 3,, ] T 3[, 3,, ] T = [, 3,, ] T = (a + b)x

3 Property m4 x = [, 3,, ] T = [, 3,, ] T = x. S 3.3: No. 5. For S = {a, d}, as given below, show that either Sp S = R or give an algebraic specification for Sp S. If Sp S R, also give a geometric specification of Sp S. ( ( a =, d = ) ) Solution: Take y = [y, y ] T R and inorder to describe the span of S set ( )( ) ( ) x y x a + x d = y implying = Augmented matrix of the above linear system on row reduction as shown below gives ( ) ( ) ( ) y y y R+R R R y y + y y + y It is clear that there is no restriction for y = [y, y ] T so that the linear system can be solved for [x, x ] T. Hence Sp S = R in this case. S 3.3: No. 8. For S = {v, w, z}, as given below, show that either SpS = R 3 or give an algebraic specification for Sp S. If Sp S R 3, also give a geometric specification of Sp S. v =, w =, z = x Solution: Take y = [y, y, y 3 ] T R 3. We seek x = [x, x, x 3 ] T so that x v + x w + x 3 z = y. This implies x y x = y x 3 y 3 Augmented matrix of the above linear system on row reduction as shown below gives y y y y R R y y R y + y y 3 y 3 y 3 y R3 R y + y y 3 + y y Since the above corresponds to a linear system for x that only has solution if y is restricted with y 3 + y y =, it is clear that we need the Sp S = { y R 3 : y 3 + y y = }, which is the equation of a plane normal to the y 3

4 vector (,, ) T. Alternately, if we take y = t, y = s, arbitrary, we have y 3 = t s and so y = [t, s, t s] T = t[,, ] T + s[,, ] T and so Sp S = Sp { [,, ] T, [,, ] } T = Sp {v, w}. S 3.3: No. 36. Same question as No. 3, but with A = Solution: As in problem 3 above, we set Ax = y, now with both x = [x, x, x 3 ] T, y = [y, y, y 3 ] T R 3, in order that matrix multiplication makes sense. Then, row reduction of the linear system leads to y y y y R+R R 3 R y + y R3 R y + y y 3 3 y 3 y y 3 3y y y y + y 3 y R R3 3y + 4y y 3 R+R3 3y + 4y y 3 y 3 3y y y 3 3y y Clearly, the system of equations is consistent regardless of y and so R(A) = R 3. Also, if we set y = above the solution set for x is just x = = [,, ] T and hence N(A) = { R 3}. (Note we are denoting R 3 since the zero vector has a different number of component zeros for different R n ). S 3.3: No. 38 For A as given below, determine for b = [, 3] T, if i. b R(A). ii. If b R(A), exhibit a vector x R such that Ax = b. iii. If b R(A), then write b as a linear combination of columns of A. ( ) A = 3 6 Solution: We seek to solve Ax = [, 3] T which gives rise to the augmented matrix below, which on row reduction gives ( ) ( ) R+3R which is consistent and has solution x = + t, x = t, arbitrary. Therefore, b = [, 3] T R(A). To answer ii, we note that x = [x, x ] T = [ + t, t] T = [, ] T +t[, ] T. So, in particular with choice t =, we have x = [, ] T satisfying Ax = b as also follows from inspection. To answer part iii. With x = [, ] T choice, we have clearly x A + x A = Ax = [, 3] T and we therefore have b = A, which is clear from inspection of the first column of matrix A and the vector b. S 3.3, No. 4. Same question as (38.) except now A is as given below A =

5 and different cases of b as listed below 4 (i)b =, (ii)b =, (iii)b = 7, (iv)b =, (v)b =, (vi)b = Solution: To answer the question for a whole lot of different b, it is better to keep b = [b, b, b 3 ] T and determine conditions on b and then check if those conditions hold for each of the b provided. Now, if we set Ax = b, it is clear that x = [x, x, x 3 ] T R 3 and the augmented matrix for the resulting linear system row reduces as shown: b b [A b] = 3 5 b R R R 3 R 3 b b 7 b 3 6 b 3 b b 7 3b + b 3 b b R+R 3 b b b 3 + 3b b b 3 + 3b b R3 R It is clear that b R(A) if and only if b 3 +3b b =. With given b, we readily check this is satisfied for cases (ii), (iii), (iv), (vi), in which case b R(A), but the condition is not satisfied in cases (i), (v). x in case (ii) may be chosen (with x 3 = ) to be [ 3b +b, b b, ] T = [ 3+,, ] T = [,, ] T and we can readily verify on inspection that indeed A A = b. Now consider case (iii) which for x 3 = leads to x = [ 3(4) + (7), 7 (4), ] T = [, ] T and we may readily verify that A A = b in this case. Now consider case (iv). We have for x 3 =, x = [ 3b + b, b b, ] T = [ 3() +, ()] T = [, ] T and accordingly we can readily verify that A + A = b. The last case, case (vi) is trivial since b = [,, ] T = A + A + A 3. S 3.3: No. 4 For W as below exhibit a 3 3 matrix A so that W = R(A) and conclude that W is a subspace. W = { x = [x, x, x 3 ] T : 3x 4x + x 3 = } We note that the description of W, implies x = s and x 3 = t are arbitrary and thus, x = 4 3 s 3 t. Therefore, for x W, we have x = [x, x, x 3 ] T = s [ 4 3,, ] T + t [ 3,, ] T and so W = Sp { [4 ] T 3,,, [ 3 ] } T,,. and therefore we may choose A = 5

6 Note R(A) is the column space of A, which is the span of the column vectors. Since the zero vector in the third column is not independent of the first two columns, the span of columns of A is W and therefore equal to R(A). Since R(A) is a subspace, it follows that W is a subspace of R 3. S 3.3: No. 44 Let S = {v, w, x} as given below is a set of vectors. Exhibit a matrix A such that Sp S = R(A). v =, w =,, x = Solution: From Matrix multiplication we know that for a 3 3 matrix A and x = [x, x, x 3 ] T, Ax = x A + x A + x 3 A 3 ; therefore, span of the column vectors of A is the same as the range of A. Therefore, choosing A = v, A = w and A 3 = x will ensure that R(A) = Sp S, and so A = 6