MATH 218 Problems Solution Manual Chapter 4.2, Fall 2011

94 Chapter 4 Linear Transformations is a linear transformation. The null space of T', denoted by N(T), is the null space of the matrix, N (A) = {x €R®| Ax = 0}. Since ; 2y 13.0 zy 1 r 83 07 T XQ =/]2 0 8 % |} =a | 2] +a f 0 | +23] 3], 3 2 0 23 2 0 3 the range of T, denoted by R(T) is the column space of A, col(A), Since’ 1 3 0 1 3 0 2 0 38 | reducesto| 0 —6 3 20 3 0 0 0 w the homogeneous equation Ax = 0 has infinitely many solutions given by a, = — 323,02 = $03, and 23 ~3/2 a free variable, So the null space is ¢ ¢] 1/2 . 1 in three space. Also since the pivots in the reduced matrix are in columns one and two, a basis for the te x} , which is a line that passes through the. origin 2 0 . 3 = dim(R*) = dim(R(L)) + dim(N(T)). This is a fundamental theorem that if 7: V —» W is a linear transformation defined on finite dimensional vector spaces, then dim(V) = dim(R(T)) + dim(N(T)).. If the mapping is given as a matrix product T(v) = Av such that A isamxn matrix, then this result is written as 1 3 range is 2 | , 0 | and hence, the range is a plane in three space. Notice that inthis example, n = rank(A) + nullity(A), A number of useful statements are added to the list of equivalences concerning n x 7 linear systems: A is invertible < Ax = b has a unique solution for every b & Ax = 0 has only the trivial solution < A is row equivalent to I det(A) #0 the column vectors of A are linearly independent « the column vectors of A span R” © the column vectors of A are a basis for R” rank(A) =n R(A) = col(A) = R" & N(A) = {0} & row(A) = R” <> the number of pivot columns in the row echelon form of A is n. || Solutions to Exercises 1. Since T(v) = , »visin N(T), 2. Since T(v) = ’ »Visin N(T), —5 . 3, Since T(v) = 10 » Vv is not in N(T), 4. Since T(v) = | , Vis in N(T). 5. Since p'(z) = 2x - 3 and p’(x) = 2, then 6, Since p'(w) = 5 and p!"(x) = 0, then T(p(x)) = T(p(z)) = 2a, so p(x) is not in N(T). 0, so p(x) is in N(T). 7. Since T(p(«)) = —2x, then p(x) is not in 8, Since T(p(x)) = 0, then p(z) is in N(T). N(T). 1 0 2/1 10 2 {1 9. Since | 2 1 8/8 | reducesto} 0 1 —1/1 | there are infinitely many vectors that are mapped 1 -1 3/0 00 0 f0 . ; 1 -1 1 | 1 to | 3 |. For example, T 2 = 38 | and hence, | 3 | is in R(T). 0 1 0 0 4.2 The Null Space and Range 95 1 0 2/2 1 0 2 40 10, Since | 2 1 3) 3.] reducesto| 0 1. ~1 the linear system is inconsistent, so the vector Lo -1 3/4 00 0j1 : [ 2 3 | is not in R(T), L 4 1 0 2) +1 1 0 240 11. Since | 2 1 38) 1 | reducesto| 0 1 —1] 0 |, the linear system is inconsistent, so the vector 1 -1 3 |-2 00 O0/}1 [ —1 1 | is not in R(T). | -2 1 0 2) -2 10 2/2 12. Since | 2 1 3/5 | reducestg| 0 1 -—1]-—1 ] there are infinitely many vectors that are 1 -1 3]~-1 00 O07 0 - —2 mapped to | ~—5 | and hence, the vector | ~5 | isin R(T). | E 13, The matrix A is in R(T), 14, The matrix A is not in R(T). 15. The matrix A is not in R(T). 16, The matrix A is in R(T). 17. A vector v = | ) is in the null‘space, if and only if 8a -+y = 0 and y = 0. That is, N(T) = {| ; \} . Hence, the null space has dimension 0, so does not. have a basis. -a+y =0 ay =0" that is = y. Therefore, N(T) Hl 18. A vector is in the null space if and only if {| ‘ ar} and hence, a basis is {| : |}. L+ 2% 0 19, Since | 2a-+-y+3z | = | 0 | if and only if x = —2z and y = z every vector in the null space has the w-—y+3z 0 : —2z —2 form z | . Hence, a basis for the null space is 1 t | f—-2 2 2 1 0 —1/2 1/2 7 20. Since | 3 5 1 | reduces to 0 1 1/2 | , then N(T) = {: | —1/2 | te x} and a basis for lo 21 00 0 1 1/2] the null space is —1/2 1 Is +t 3 21. Since N(T) = 8 s,t€R>,a 22. A basis for the null space is 1 . t 0 (f 2 1 basis for the null space is | 1 | ; 0 | . 0 1 96 Chapter 4 Linear Transformations 23, Since T(p(x)) = 0 if and only if p(0)=Oa 24. If p(x) = ax? + ba +e, then p'(z) = 2ax +b polynomial is in the null space if and only if it and p’(x) = 2a, so T(p(x)) = 0 if and only if - has the form ax? + bx. A basis for the null space a = 0. A basis for the null space is {1,2}. is {v, 27}. 1 1 2 25. Since det (| 04 -1 } = —6, the column vectors of the matrix are a basis for the column 20 1 space of the matrix. Since the column space of the matrix is R(T), then a basis for the range of T is thE . 1 -2 -3 1 5 1010 1 26. Since | 3 -1 1 O 4 | reducestg| 0 1 2 0 -1 | and the pivots are in columns one, two, 1 ot 38 1 2 000 1 2 —2 1 and four, then a basis for the column space of A. and hence, for R(T), is | 3 | , | -1 | , 0 | . : 1 1 1 1 0 27. Since the range of Tis the zy-plane in R°, a basis for the range is | 0 | ; | 1 | . ; “TO 0 w-y+3z 1 —1 3 28, Since atytz se @ | -L +y}] ol +2] 1 and the three vectors are linearly inde- : : —2 + 8y — Bz 1 3 -5 : : l -1 3 pendent, then a basis for R(T’) is 1 , 1 ; 1 ~1 3 -5 29, Since R(T) = Pg, then a basis for the range 30. Since is {1,2,27}, R(T) = {p(x) | p(w) = aa*-+-ba+a = a(2?+-1)-+b2}, then a basis for R(T) is {a, 2? + 1}. 31. a, The vector w is in the range of T if the linear system —2 0 —2 6 Cy 1 + eg 1 +3 2 = 5 1 -1 0 0 -2 0 -2|-6 —-2 0 -2 has a solution, But 1 1 27 5 — Oo 1 ii ~6 2 | , so that the linear system is inconsis- 1 -1 0] 0 0 0 0 }-1 1 6 i tent. Hence, | 5 | is not in-R(t). 0 —~2 0 -2 b. Since | ‘1 1 2 |= 0, the column vectors are linearly dependent. To trim the vectors to a basis for 1 -1 0 -—2 0 —2 -2 0 -2 the range, we have that 1 1 2 oS 0 tii . Since the pivots are in columns one and 1-1 0 0 0 0 4.2 The Null Space and Range ‘ 97 . 2 0 . two, a basis for the range is 1 | : 1 .¢. Since dim(N(T)) + dim(R(T)) = dimi(IR?) = 3 and 1 dim(R(T')) = 2, then dim(N(T)) = 1. -1 —2 —l 0 -l 32, a. The vector | 1 | isin R(T), b. 2 1,) 5],]) -1 c. Since 2 1 0 2 dim(N(T)) + dim(R(T)) = 3 and dim(R(L)) = 3, then dim(N(TL)) = 0. .-- . 33. a. The polynomial 2a? — 4x + 6 is not in R(T). b. Since the null space of 7’ is the set. of all con- stant functions, then dim(N(L)) = 1 and hence, dim(R(T)) = 2. A basis for the range is {T'(x),T(2*)} = {-2¢-+1,0? +a}. 34. a. The polynomial 2?—2—2 is not in R(T). b. Since the null space of T is the set of all polynomials of the form ax?, then dim(N(T')) = 1 and hence, dim(R(T)) = 2. A basis for the range is {7(1), T(2)} = {2,2 ~1}. 35. Any linear transformations that maps three space to the entire xy-plane will work.’ For example, the x mapping to the zy-plane is T’ {| y } = ' . z ! 86, Define TR? R?, wy T (| *])—=[ 8]. ‘Then N(L) = {[| 37. a. The range R(T’) is the subspace of P, consisting of all polynomials of degree n ~ 1 or less. b. dim(R(T)) =n — ¢. Since dim(R(L)) + dim(N(T)) = dim(Pp) = n +1, then dim(N(T)) = 1. 38, A polynomial is in the null space provided it 39. a, dim(R(T)) = 2 b, dim(N(T)) = has degree k ~ 1 or less. Hence dim(N(T)) =k 40. Since dim(V) = dim(N(T)) + dim(R(T)) = 2dim(N(T)), then the dimension of V is an even number, LE R} = R(T), op} a d “he _ _ 0 2b . f 41, If B= al? then T(B) = AB- BA= 2 9 |) that : N(T) = {[6 a adeR} = {ol 4 0 | tale || adr}, Hence a basis for N(T) is 1 0 0 0 . : 0 O0;7,0 1 : 42. If B is an n x n matrix, thon T(B*) = (B*)' = B and hence, R(T) = Maxn. 43, a. Notice that (A+A‘)! = A'+A = A+ A’, so that the range of T is a subset of the symmetric matrices. Also if B is any symmetric matrix, then T ($B) = 4B+4B' = B. Therefore, R(T) is the set of all symmetric matrices, b. Since a matrix A is in N(T) if and only if T(A) = A+ A! = 0, which is if and only if A= —A‘, then the null space of T is the set of skew-symmetric matrices. 44, a. Notice that (A+ A‘)! = A'—A = —(A—A‘), so that the range of ri is a subset of the skew-symmetric matrices. Also if B is any skew-symmetric matrix, then T (3B) = 4B —4B! = B. Therefore, R(T) is the set of all skew-symmetric matrices. b. Since a matrix A is in N(T) if and only if T(A) = A—A® = 0, which is if and only if A-= A‘, then the null space of T is the set of symmetric matrices, 45, Ifthe matrix A is invertible and B is any nxn matrix, then T(A7!B) = A(A7!B) = B,so R(T) = Mnxn- 46, a, A basis for the range of T consists of the column vectors of A corresponding to the pivot columns of the echelon form of A, Any zero rows of A correspond to diagonal entries that are 0, so the echelon form of A will have pivot columns corresponding to each nonzero diagonal term, Hence, the range of T is spanned by the nonzero column vectors of A and the number of nonzero vectors equals the number of pivot columns of A. b, Since dim(N(T)) = n = dim(R(T)), then the dimension of the null space of T is the number of zeros on the diagonal. i i i i