# YORK UNIVERSITY. Faculty of Science Department of Mathematics and Statistics MATH M Test #1. July 11, 2013 Solutions

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1 YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH M Test # July, 23 Solutions. For each statement indicate whether it is always TRUE or sometimes FALSE. Note: For this question only, each correct answer is worth.5 point and each incorrect answer is worth.5 (negative half!) point. If the number of incorrect answers is more than three times 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, you do NOT need to explain your answer or show your work. Statement TRUE/FALSE If U is a subspace of R n and X U, then ( X) U. The set of all invertible 3 3 matrices is a subspace of M 33. If span A = span B, then there are the same number of vectors in A and B. If {X, Y, } is linearly dependent, then is {X, Y, Z} is also linearly dependent. If U = span{x, Y } and V = span{x, X + Y }, then dim U = dim V. If {X, X 2, X 3, X 4 } and {Y, Y 2, Y 3, Y 4 } are bases of R 4, then {X + Y, X 2 + Y 2, X 3 + Y 3, X 4 + Y 4 } is also a basis of R 4. Similar matrices have the same eigenvalues. If R is the reduced row echelon form of a matrix A, then A is similar to R.

2 2. In each case determine whether U is a subspace of R 3. If so, find its dimension. Justify each answer you give. (a) U = {[ s 2s] T s R}. U is a subspace of R 3, because it contains a zero vector, and it is closed under addition and scalar multiplication. dim U =. (b) U = {[s t s + t] T s, t R}. U is a subspace of R 3, since it contains a zero vector, and it is closed under addition and scalar multiplication. dim U = 2. (c) U = {[s s + t t] T s, t R}. U is not a subspace, because e.g., it does not contain a zero vector. Note that U is not closed under addition or scalar multiplication. 3. Find a basis and calculate the dimension of the following subspace of R 4 U = span{[ 3 ] T, [ 2] T, [ 2 4 ] T, [ 2 ] T }. Writing the spanning vectors as columns, we obtain the matrix A = So, U = col A. The row-echelon form of the matrix A is /2 /2 Hence, columns, 2 and 4 of A will form a basis for col A and consequently, for U. Hence, the set {[ 3 ] T, [ 2] T, [ 2 ] T } will be a basis for U. Counting the number of vectors in the basis, we conclude that dim U = 3. Note that the problem can also be solved using a row space. 2

3 4. (a) Show that span{[ ] T, [ ] T, [ ] T } = R 3. Denote R = [ ], R 2 = [ ] and R 3 = [ ] and consider the matrix A = [R R 2 R 3 ] T. Then a row-echelon form of the matrix A, has 3 nonzero rows. Hence, {R, R 2, R 3 } is linearly independent. Therefore, span{[ ] T, [ ] T, [ ] T } = row A = R 3. (b) Determine whether the vector 3x 2 is in span{3x +, x 2 + 2x + 3}. Justify your answer. If vector 3x 2 is in span of S, then there must exist scalars s and t such that 3x 2 = s(3x + ) + t(x 2 + 2x + 3) = tx 2 + (2t + 3s)x + (3t + s). Equating the coefficients of the corresponding terms, we obtain the following system of linear equations: t = 3, 2t + 3s =, 3t + s =. Substituting t = 3 in the 2nd equation, we obtain s = 2 and substituting t = 3 in the 3rd equation, we obtain s = 9. Hence, the system is inconsistent and therefore, 3x 2 is not in span of S. [ ] [ ] [ ] (c) Determine whether the set of vectors {,, } is linearly independent. Do they form a basis for M 22? Justify your answer. Let [ ] [ ] a + b = Then we obtain the following homogeneous system of four linear equations with two unknowns: a + 2b =, 3a + 2b [ =, a ] 3b = [, 2a + ] 4b =, which has only the trivial solution a = b =. So, the set {, } is linearly independent [ ] [ ] [ ] On the other hand, is not in span {, } Indeed, if [ ] [ ] [ ] = r + s, we obtain the following inconsistent system: r + 2s = 2, 3r + 2s = 3, r 3s =, 2r + 4s = 2. Therefore, the original set is linearly independent. Finally, since dim M 22 = 4, the set of three vector is not a basis for M (a) Let X and Y be nonzero vectors in R n. Show that X + Y 2 = X 2 + Y 2 if and only if X is orthogonal to Y. X + Y 2 = (X + Y ) (X + Y ) = X X + 2X Y + Y Y = X 2 + 2X Y + Y 2. 3

4 So, X + Y 2 = X 2 + Y 2 2X Y = X Y = X Y. (b) Let {F, F 2,, F n } be an orthogonal basis of R n. Then for every X R n, which of the following statements is true: i. F i F j, i j; ii. X 2 = ( X F F )2 + ( X F 2 F 2 )2 + + ( X Fn F n )2 ; iii. X = (X F )F + (X F 2 )F (X F n )F n ; iv. F + F F n = ; v. None of the above-mentioned? Explain. (ii) is true. Since {F, F 2,, F n } is an orthogonal basis of R n, X R n, X = X F F 2 F + X F 2 F 2 2 F X F n F n 2 F n. So, X 2 = X X = ( X F F 2 F + X F 2 F 2 2 F X F n F n 2 F n) ( X F F 2 F + X F 2 F 2 2 F X F n F n 2 F n) = ( X F F )2 + ( X F 2 F 2 )2 + + ( X F n F n )2 F i F j =, i j and F i F i = F i 2, i =, 2,, n. 6. Consider the matrix M = (a) Find the basis for the nullspace of the matrix M. null M = {X R 4 MX = }. The row-echelon form of the matrix M is /7. So, the solution to MX = will be x 4 = t, x 3 = s, x 2 = x x 4 = s 4 7 t, 4

5 or in the matrix form Hence, and consequently, the set will be a basis for null M. x = 4x 2 5x 3 2x 4 = s t, X = s null M = {X R 4 X = s + t + t 2/7 4/7 2/7 4/7 {[ ] T, [2 4 7] T } (b) Calculate rank M directly from your answer to part (a). So, dim (null M) = n rank M. rank M = n dim (null M)., s, t R}, But from part (a), dim (null M) = 2. Hence, rank M = n 2 = 4 2 = 2. Or equivalently, counting numbers of nonzero rows in a row-echelon form of matrix M. 7. Given the matrix A = Find matrices P and D such that P AP = D, where D is a diagonal matrix. C(x) = det (xi 3 A) = det x x x = (x ) 3 (x ) = (x )[(x ) 2 )] = (x )(x 2 2x) = x(x )(x 2). So, the eigenvalues λ =, λ 2 = and λ 3 = 2. Solving the corresponding linear systems AX = λ i X, i =, 2, 3; we find the eigenvectors. The eigenvector corresponding to λ = is X = [ ] T, the eigenvector corresponding to λ 2 = is X 2 = [ ] T and the eigenvector corresponding to λ 3 = 2 is X 3 = [ ] T. Hence, the diagonalizing matrix P = [X X 2 X 3 ] = 5.

6 and consequently, that is P AP = diag (,, 2), D = 2. The end. 6

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