YORK UNIVERSITY. Faculty of Science Department of Mathematics and Statistics MATH M Test #2 Solutions

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1 YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH 3. M Test # Solutions. (8 pts) For each statement indicate whether it is always TRUE or sometimes FALSE. Note: For this question each correct answer is worth point and each incorrect answer is worth.5 (negative half!) points. If the number of incorrect answers is more than twice greater than the number of correct ones, then the total mark will be zero. If you don t know the answer, don t write anything. For this question only, you do NOT need to explain your answer or show your work. Statement TRUE/FALSE If {u,v,w} is linearly independent, then so is {u+w,v+w}. The set of all invertible matrices is a subspace of M. False If U is a subspace of a vector space V, then dim U < dim V. False Both vector spaces P 5 and M 3 are isomorphic to R 6. For any matrix A, both A T A and AA T are orthogonally diagonalizable. If T : V W is a linear transformation and for some vector v V, T ( v) = T (v), then v =. False Let T : R 3 R 3 be a linear operator. Then if im T is a plane through the origin in R 3, then ker T is a line through the origin in R 3. If T : V W is an isomorphism, then there exists an isomorphism S : W V.

2 . (6 pts) Find the least squares approximating line for the following points: From the data, M T M = x x x 3 x 4 (, ), (, 3), (, ), (3, 6). M T Y = x x x 3 x 4 x x x 3 x 4 = y y y 3 y 4 = = = 3 So, the associates system of normal equations (M T M)Z = M T Y for Z = z z T will be 4 4 z3 = 4 4 z Solving it using Gaussian elimination, we obtain z = 5 and z = 7. Therefore, the least squares approximating line for the given points is y = z + z x = x. 3. (4 + 4 pts) a b a (a) Let U be the set of all matrices of the form c a subspace of M? If so, find its basis, and determine dim U. For all a, b, c R, a b a c a = a So, U = span {,, M, we conclude that U a subspace of M. On the other hand, {,, r + s + b, where a, b, c R. Is U a + c. }. Since the spanning set is a subset of } is linearly independent because + t = implies that r = s = t =. Therefore, the set { dim U = 3.,, } is a basis for U and consequently,

3 (b) Let W = {f(x) P 4 f(x) = ax 4 + bx, where a, b R}. Is W a subspace of P 4? If so, find its basis, and determine dim W. W =span{x, x 4 } implies that w is a vector space and since the spanning set is a subset of P 4, W is a subspace of P 4. On the other hand, {x, x 4 } is linearly independent because rx + sx 4 = implies that r = s =. Therefore, {x, x 4 } is a basis for W and dim W =. 4. (3 + 4 pts) (a) Determine whether T : P 3 M defined by T (a + bx + cx + dx 3 ) = is a linear transformation. Justify your answer. = 4 Let a = c = d = and b =. Then T (x) = while T (x) = =. Hence, T (x) T (x) and Axiom is not satisfied for r =. Therefore, T is not a linear transformation. (b) Let T : R 3 M be a linear transformation defined by T (a, b, c) = Then which of the following sets is a basis for im T? i. {, } ii. { } iii. {, } iv. {,, } v. none of the above-mentioned., 3c a + 4d a + b + c + d (b + c) a a + b b + c c Explain. (iv). a b + c If T (a, b, c) =, then =. Solving the corresponding system, we obtain a = b = c =. So, ker T = {} and dim (ker T ) =. a + b c By the Dimension Theorem, dim (R 3 ) = dim (ker T)+ dim (im T). Hence, dim (im T)= 3 = 3. Since set iv is the set of three linearly independent vectors, it is a basis for im T.. 5. (8 + + pts) Let T : P P be a linear transformation defined by (a) Determine: T (c + c x + c x ) = c + c (x + ) + c (x + ). 3

4 i. a basis for im T ; ii. a basis for ker T ; iii. rank (T ) and nullity (T ). If T (c + c x + c x ) =, then c + c (x + ) + c (x + ) =, that is (c + c + c ) + (c + 4c )x + 4c x =. Equating the coefficients to zero, we obtain a homogeneous linear system which has only the trivial solution c = c = c =. So, ker T = {} and consequently, nullity T = dim (ker T )= and a basis of ker T =. By the dimension Theorem, dim (P ) = dim (ker T )+ dim (im T ). So, rank T = dim (im T ) = 3 = 3 and im T = P, and its basis will be {, x, x }. (b) Is T one-to-one? Explain. From part (a), ker T = {}. So, by Theorem (Section 7.) T is one-to-one. (c) Is T onto? Explain. From part (a), im T = P. So, T is onto. 6. ( pts) (a) Which of the following can be added to {x 5, x 3x 5} to form a basis for P 3? i. {7x, x 3 } ii. {, x, x } iii. {4x 3 + x, x 3 } iv. {5x 3, x 3 } v. none of the above-mentioned. Explain. (iii). dim (P 3 ) = 4. t (x 5)+t (x 3x 5)+t 3 (4x 3 +x )+t 4 (x 3 ) = implies (t t )5 3t x (t +t + t 3 )x + (4t 3 + t 4 )x 3 =. Equating the coefficients to zero, we obtain a homogeneous linear system which has only the trivial solution t = t = t 3 = t 4 =. Hence, {x 5, x 3x 5, 4x 3 + x, x 3 } is a set of four linearly independent vectors that forms a basis for P 3. (b) Let T : V W be a linear transformation, where dim V = 5. Then which of the following statements is true? i. If T is one-to-one, then dim (im T ) dim (ker T ) ii. If dim (ker T ) =, then dim (im T ) = 3 iii. If T is onto, then dim (ker T ) iv. T = V v. none of the above-mentioned. Explain. (ii). By the Dimension Theorem, for linear transformation T : V W, 4

5 dim (V ) = dim (ker T )+ dim (im T ). So, dim V = 5 and dim (ker T ) =, then dim (im T ) = dim (V ) dim (ker T ) = 5 = 3. (c) Show that {A, A,, A k } is a basis for M mn if and only if {A T, AT,, AT k } is a basis for M nm. Let T : M mn M nm be a linear transformation defined by T (A) = A T, A M mn. Then T (A) =, i.e. a zero matrix of n m = A =, i.e. a zero matrix of m n. Hence, ker T = {} and consequently, a basis for ker T is. On the other hand, dim (M mn ) = mn = nm = dim (M nm ). Therefore, if {A, A,, A k } is a basis for M mn and is a basis for ker T, then k = mn and by Theorem 5 (Section 7.), {T (A ), T (A ),, T (A k )} = {A T, AT,, AT k } will be a basis for im T = M nm. Similarly, since (A T ) T = A, applying Theorem 5 (Section 7.) to a linear transformation S : M nm M mn defined by S(A T ) = (A T ) T = A, A T M nm, we obtain that if {A T, AT,, AT k } is a basis for M nm, then {A, A,, A k } is a basis for M mn. 7. (6 pts) Express the matrix B = as a product QR, where Q is an orthogonal matrix and R is an upper triangular matrix. Write B = C C. Then applying the Gram-Schmidt algorithm, and normalizing we obtain Hence, = = 8. (8 pts) Consider the matrix F = C = T, F = C C F F F = 3 3 T, Q = F F = T, Q = F F = T. F C B = QR = Q Q Q F 3 T 3 A = = 3 Find an orthogonal matrix P and a diagonal matrix D such that P AP = D. 5.

6 C A (x) = det(xi 3 A) = det x x x So, eigenvalues of A are: λ = and λ = (multiplicity of ). The eigenvector corresponding to λ = is X =. For λ =, the general solution to the corresponding system will be X = s + t ( s, t R). So, we obtain two more eigenvectors X = and X 3 = Please note that the set {X, X, X 3 } is orthogonal. Hence, normalizing, we obtain P = X X =, P = X X =. = x (x ). and P 3 = X 3 X 3 =. Thus, P = P P P 3 = =. Note that the columns of P are orthonormal and P AP = P T AP = D = diag(,, ). The end. 6