2 b 3 b 4. c c 2 c 3 c 4

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1 OHSx XM511 Linear Algebra: Multiple Choice Questions for Chapter 4 a a 2 a 3 a 4 b b 1. What is the determinant of 2 b 3 b 4 c c 2 c 3 c 4? d d 2 d 3 d 4 (a) abcd (b) abcd(a b)(b c)(c d)(d a) (c) abcd(a b)(a c)(a d)(b c)(b d)(c d) (d) 0 Solution: (c). Although it is possible to directly use cofactor expansion along a column or row, for matrices of size 4 4 or greater, the calculations often become unwieldy. In such cases, it is often most efficient to use elementary row operations to row reduce the matrix and keep track of the effect on the determinant. a a 2 a 3 a 4 1 a a 2 a 3 b b det 2 b 3 b 4 c c 2 c 3 c 4 = (abcd)det 1 b b 2 b 3 1 c c 2 c 3 d d 2 d 3 d 4 1 d d 2 d 3 = (abcd)det 1 a a 2 a 3 0 b a b 2 a 2 b 3 a 3 0 c a c 2 a 2 c 3 a 3 0 d a d 2 a 2 d 3 a 3 = (abcd)(b a)(c a)(d a)det = (abcd)(b a)(c a)(d a)det = (abcd)det 1 b + a b 2 + ab + a 2 1 c + a c 2 + ac + a 2 1 d + a d 2 + ad + a 2 b a b 2 a 2 b 3 a 3 c a c 2 a 2 c 3 a 3 d a d 2 a 2 d 3 a 3 1 b + a b 2 + ab + a 2 0 c b c 2 b 2 + ac ab 0 d b d 2 b 2 + ad ab c b c = (abcd)(b a)(c a)(d a)det 2 b 2 + ac ab d b d 2 b 2 + ad ab ( ) = (abcd)(b a)(c a)(d a) (c b)(d 2 b 2 + ad ab) (d b)(c 2 b 2 + ac ab) = (abcd)(b a)(c a)(d a) ((c b)(d b)(d + b + a) (d b)(c b)(c + b + a)) = (abcd)(b a)(c a)(d a)(c b)(d b)(d c). In general, a matrix of the form 1 λ 1 λ 2 1 λ n λ 2 λ 2 2 λ n λ n 1 λ 2 n 1 λn 1 n 1 is called a Vandermonde matrix. Its determinant is (λ i λ j ). i<j 1

2 2. Suppose det (a) det (b) det (c) det (d) det = 5 a b c 2d 2e 2f g h i = 5. Which is false? = 10 d 3a e 3b f 3c a + g b + h c + i a g d b h e c i f = 5 = 5 Solution: (d). The key is to know the effect of row operations and taking the transpose on the determinant, and to figure out which operations convert the given matrices to the original (or vice-versa). (a): det = ( 1)det = ( 1) 2 det (b): (c): det det a b c 2d 2e 2f g h i d 3b e 3b f 3c a + g b + h c + i = det = ( 1)2( 1)det a + g b + h c + i = det (d): det a g d b h e c i f = det = ( 1)det 3. Let A be an n n matrix. Which condition is not equivalent to the others? (a) If Av = Aw then v = w. (b) For every v R n there exists u R n such that Au = v. (c) ker(a) = {0}. (d) det(a) = 0. 2

3 Solution: (d). Condition (a) is that A is one-to-one and this is equivalent to (c) (Lecture 29). Condition (b) is that A is onto, which is equivalent to the condition that r(a) = n. By the rank-nullity theorem, this is equivalent to condition (c). On the other hand, (d) is equivalent to A not being one-to-one. 4. Let A be an n n matrix and C the n n matrix of its cofactors: Recall from Lecture 30 that the minor µ ij (A) is the (n 1) (n 1) submatrix of A obtained by deleting the ith row and jth column, and that the cofactor c ij = ( 1) i+j det(µ ij (A)). Which describes the diagonal entries of AC t? (The matrix C t is called the adjugate of A.) (a) All the diagonal entries are zero. (b) The (1, 1) entry is det(a), and the other diagonal entries are all zero. (c) The (1, 1) entry is det(a), but the other diagonal entries do not have a simple description. (d) Every diagonal entry equals det(a). Solution: (d). Recall that the determinant of A can be obtained by cofactor expansion along any row or column. Consider the (i, i) entry b ii of AC t. Since the (k, i) entry of C t is c ik, it follows from matrix multiplication that b ii = n a ik c ik = k=1 n a ik ( 1) i+k det(µ ik (A)). This is none other than the cofactor expansion of A along the ith row. b ii = det(a) for all i = 1,..., n. k=1 Therefore, We can also find the non-diagonal entries. Using an observation from Lecture 31, it s not hard to see that if i j, the (i, j) entry b ij of AC t is 0. Note that b ij = n a ik c jk = k=1 n a ik ( 1) j+k det(µ jk (A)). k=1 As observed in Lecture 31, the quantity on the right is the determinant of the matrix A obtained by replacing the jth row of A with a copy of its ith row. But since the ith and jth rows of A are identical (and i j), it follows that det(a ) = 0. Thus, b ij = 0 for i j, as claimed. Notice from the above that AC t = det(a)i. Thus if det(a) 0, that is, if A is invertible, then A 1 = 1 det(a) Ct. 5. Suppose A and B are similar matrices. Which need not be the same? (a) Their characteristic polynomials. (b) Their determinants. (c) Their kernels. (d) Their ranks. 3

4 Solution: (c). (a): See Lecture 35. (b): See Lecture 32. (c): If B = P AP 1 for some invertible matrix P, then it is easy to see that Bx = 0 if and only if A(P 1 x) = 0. So their kernels may not be the same. For example, if and P 1 = then AP t = A0 1 t = 0, so B1 0 t = 0, but A1 0 t 0. (d): By the first remark on (c), it follows that the nullities of A and B are equal. Therefore, their ranks are equal. (Or more generally, multiplying an invertible matrix on the left or right does not change the rank of a matrix, see Problem 25 in Chapter 2.) 6. Suppose A and B are square matrices of the same size. Which is always true? (a) det(a B) = ±det(b A) (b) If det(a) = det(b), then det(a B) = 0. (c) If ker(a) = ker(b) then image(a) = image(b). (d) If image(a) = image(b) then ker(a) = ker(b). Solution: (a). (a): If A and B have size n n, then det(a B) = ( 1) n det(b A). (b): Obviously not true. For example, consider when B = I. Then det(a) = det(b) = 1, but det(a B) = 0. (c): If ker(a) = ker(b), then the nullities of A and B are equal, so the the dimension of their images are equal, but not necessarily the images themselves. For example, if and B = then ker(a) = ker(b) = span{e 1 }. However, image(a) = span{e 1 }, whereas image(b) = span{e 2 }. (d): Similarly, if image(a) = image(b) then their ranks are the same, so the dimension of their kernels are the same, but not necessarily the kernels themselves. For example, if and B = then their images both equal span{1 1 t }, but ker(a) = span{0 1 t } and ker(b) = span{ 1 1 t }. 7. Suppose A has characteristic polynomial (x 1) 2. Which need not be true? (a) A is a 2 2 matrix. (b) A has at least one eigenvector. (c) A is the identity matrix. (d) A is invertible.,,, 4

5 Solution: (c). (a): A square matrix has size n n if and only if its characteristic polynomial has degree n. (b): Since A has an eigenvalue, namely 1, it has at least one eigenvector. (c): This is not necessarily true, since for example, has characteristic polynomial (x 1) 2. A comprehensive answer can be given by writing A as a b. c d The characteristic polynomial of this matrix is x 2 (a + d)x + (ad bc). Then the hypothesis is equivalent to the conditions that a + d = 2 and ad bc = 1. There are infinitely many 4 tuples (a, b, c, d) that satisfy these two equations. (d): Using the same notation as in the remark on (c), we see that since det(a) = ad bc = 1, it follows that A is invertible. 8. Which procedure is incorrect? (a) Calculating the determinant of a large matrix A: Perform Gaussian elimination to make it into an upper-triangular matrix A, keeping track of only those row operations that switch two rows or multiply a row by a scalar. Let k be number of switches and c 1,..., c n the scalar multiples. Then det(a) = ( 1) k c 1... c n det(a ). (b) Calculating the determinant of triangular matrices: The determinant of an uppertriangular or lower triangular matrix is the product of the diagonal entries. (c) Finding eigenvalues: Calculate the characteristic polynomial p(λ) of A. It is possible that the equation p(λ) = 0 has no solutions (in R). In this case A has no eigenvalues. Otherwise, the solutions are the eigenvalues. (d) Finding eigenvectors: If λ is an eigenvalue of A, then the corresponding eigenvectors are the non-zero solutions to (A λi)x = 0. Solution: (a). (a): The correct relation is that det(a) = ( 1) k c c 1 n det(a ). (See Lecture 31.) (b): See Lecture 30. (c): By definition, the (real) eigenvalues are the solutions to p(λ) = 0 in R. It can certainly happen that A has no real eigenvalues, for example, when (d): This is the definition of eigenvector. 9. If λ 1,..., λ n are distinct eigenvalues of a linear transformation T : V V, and v j is any eigenvector corresponding to λ j, for j = 1,..., n, then v 1,..., v n are linearly independent. The following gives a proof. We proceed by induction on the number n of distinct eigenvalues. (i) First, suppose n = 1. Then since v 1 0, the statement is true in this case. 5

6 (ii) Assuming that the statement is true for n = k, we will show that it is true for n = k + 1. Suppose c 1 v c k+1 v k+1 = 0 for some scalars c 1,..., c k+1. On the one hand, applying T to both sides of the equation gives T (c 1 v c k+1 v k+1 ) = c 1 λ 1 T v c k+1 λ k+1 T v k+1 = 0. ( ) On the other hand, multiplying λ k+1 to both sides gives c 1 λ k+1 v c k+1 λ k+1 v k+1 = 0. ( ) Subtracting this from ( ) we get c 1 (λ 1 λ k+1 )v c k (λ k λ k+1 )v k = 0. It follows from the induction hypothesis that since, the vectors v 1,..., v k are linearly independent. And since, it follows that c 1 = = c k = 0. Since c 1 v c k+1 v k+1 = 0 and, it follows that c k+1 = 0 as well. Here is a list of possible answers: (I) λ j λ k+1 for j = 1,..., k (II) λ 1,..., λ k are all distinct (III) v 1,..., v k are linearly independent (IV) v 1,..., v k+1 are linearly independent (V) v k+1 0 Which correctly fills the blanks in order? (a) (I) and (II); (III); (IV) (b) (II); (I); (IV) (c) (I) and (II); (III); (V) (d) (II); (I); (V) Solution: (d). The induction hypothesis is: If λ 1,..., λ k are distinct eigenvalues of T, and T v j = λ j v j, for j = 1,..., k, then v 1,..., v k are linearly independent. Therefore, (II) correctly fills the first blank. This implies that c j (λ j λ k+1 ) = 0 for all j = 1..., k. From the hypothesis that λ j λ k+1 = 0 for all j = 1..., k, we can then conclude that c 1 =... = c k = 0. Thus, it follows from ( ) that c k+1 v k+1 = 0. Now since v k+1 0, we can conclude that c k+1 = Which is not diagonalizable? (a) (b) (c) (d)

7 Solution: (d): (a): We calculate the characteristic polynomial by expanding along the second row: 1 λ λ 2 det 3 4 λ 0 = 3 + (4 λ) 1 3 λ 3 3 λ λ = 3(12 4λ + 2) + (4 λ)(λ 2 2λ 9) = 12λ + 42 (λ 3 6λ 2 λ + 36) = λ 3 + 6λ 2 11λ + 6 = (λ 1)(λ 2)(λ 3). Since the three eigenvalues are all distinct, the matrix is diagonalizable. (By finding the eigenvectors, you can further check that for P = 1 3 3, we have P 1 AP = ) (b): In this case, it is particularly easy to calculate the characteristic polynomial since the matrix is lower-triangular: λ 0 0 det 0 λ 0 = λ 2 (1 λ) λ The eigenvalue λ = 0 has a 2 dimensional eigenspace (spanned by t and t ) had λ = 1 has a 1 dimensional eigenspace (spanned by t ). Thus the matrix is diagonalizable. (c): We calculate the characteristic polynomial by expanding along the last column: det λ λ λ λ = (1 λ)det λ λ λ = (1 λ)( λ(λ 2 + 2λ 1) ( 2)) = (1 λ)( λ 3 2λ 2 + λ + 2) = (λ 1)(λ + 1)λ 2. The eigenvalue λ = 1 has a 2 dimensional eigenspace (spanned by t and t ); the eigenvalue λ = 2 has a 1 dimensional eigenspace (spanned by t ); the eigenvalue λ = 1 has a 1 dimensional eigenspace (spanned by t ). Thus the matrix is diagonalizable. (d): Since the matrix is block-diagonal, the characteristic polynomial is: det 10 λ λ λ λ det 10 λ λ = ((λ 10)(λ + 2) + 36)((λ 2)(λ + 2) + 7) = (λ 2 8λ + 16)(λ 2 + 3) = (λ 4) 2 (λ 2 + 3). 7 det 2 λ λ

8 Hence, the only (real) eigenvalue is λ = 4, which has a 1 dimensional eigenspace (spanned by t ). Thus this matrix is not diagonalizable. 11. Let D be the diagonal matrix with distinct diagonal entries λ 1, λ 2, λ 3,..., λ 10 from upper-left to lower-right. Let C be the diagonal matrix obtained from D by interchanging the second and seventh diagonal entries. Which is true? (a) If E is the elementary matrix obtained from I by interchanging the second and seventh rows, then CE = ED. (b) If we consider the linear transformation T from R 10 to itself such that T e k = λ k e k, then D represents T with respect to the standard basis. (c) If we consider the ordered basis {v 1,..., v 10 } such that v i = e i for i 2, 7; v 2 = e 7 ; and v 7 = e 2, then the matrix representation of T with respect to the ordered basis {v 1,..., v 10 } is C. (d) All of the above. Solution: (d). (a): Since the matrix E also can be described as being obtained from the identity matrix by interchanging the second and seventh columns, multiplying E on the right of a matrix A has the effect of interchanging the second and seventh columns of A. Therefore, CE is the matrix obtained from C by interchanging the second and seventh columns of C. Specifically, for k 2, 7, the kth column of CE is λ k e k ; the second column of CE is λ 2 e 7 ; and the seventh column of CE is λ 7 e 2. On the other hand, ED is the matrix obtained from D by interchanging the second and seventh rows of D. It is easy to check that this equals CE. (b): This follows from the definition of matrix representation of a linear transformation. (c): This follows from the fact that: T v k = λ k v k for k 2, 7; T v 2 = λ 7 v 2 ; and T v 7 = λ 2 v Let T : R 3 R 3 be defined by T (x 1, x 2, x 3 ) = (2x 1 x 2 x 3, x 1 x 3, x 1 + x 2 + 2x 3 ). Which is a basis for the span of all eigenvectors of T? (a) {(1, 1, 1), (1, 0, 1)} (b) {(1, 1, 1), (1, 1, 0)} (c) {(1, 1, 1), (1, 1, 0), (1, 0, 1)} (d) {(1, 1, 1), (1, 1, 0), (1, 0, 1)} Solution: (c). With respect to the standard basis, the matrix A representing T is: The characteristic polynomial of A is 2 λ 1 1 det 1 λ λ

9 λ = (2 λ)det ( 1)det 1 2 λ 1 2 λ = (2 λ)(λ 2 2λ + 1) + ( λ + 2 1) (1 λ) = (2 λ)(λ 1) 2. + ( 1)det 1 λ 1 1 To find an eigenvector for the eigenvalue λ = 2, we row reduce A 2I to get: A 2I = Hence, {(1, 1, 1)} is a basis for this eigenspace. To find an eigenvector for the eigenvalue λ = 1, we row reduce A I to get: A I = Hence, {(1, 1, 0), (1, 0, 1)} is a basis for this eigenspace. 13. Suppose A is an n n matrix with n distinct eigenvalues. Which cannot be true? (a) ker(a) has dimension 0 (contains only the zero vector). (b) ker(a) has dimension 1. (c) ker(a) has dimension 2. (d) A is diagonalizable. Solution: (c). (a): If zero is not an eigenvalue, then A is invertible, so ker(a) has dimension 0. (b): If zero is an eigenvalue, then ker(a) has dimension at least 1. On the other hand, a collection of eigenvectors corresponding to distinct eigenvalues are linearly independent. It follows that image(a) has dimension n 1, so ker(a) has dimension 1. (c): This is not possible for the reasons given above. (d): If A has distinct eigenvalues, then the eigenvectors span all of R n and so A is diagonalizable. 14. Suppose A be a square matrix that is not invertible. Which cannot be true? (a) A is diagonalizable. (b) A is not diagonalizable. (c) A has a negative eigenvalue. (d) The eigenvalues of A are all negative. Solution: (d). (a): This is certainly possible, for example,

10 (b): This is certainly possible, for example, (c): The same example as in (a) shows this is possible. (d): Since A is not invertible, 0 must be an eigenvalue. 15. Suppose A has characteristic polynomial x(x 1)(x 2). Which is not true? (a) The eigenvectors of A span R (b) A is similar to (c) A is not invertible. (d) image(a) has dimension 1. Solution: (d). (a): Since the eigenvalues of A are all distinct, the eigenvectors span R 3. (b): Let v i be an eigenvector of A associated to i, for i = 0, 1, 2, and P the matrix whose columns are v 2, v 0, v 1, in this order. Then P DP 1, so A is similar to D. (c): Since zero is an eigenvalue, A is not invertible. (d) Using the notation of the remark on (b), we have image(a) = span{v 1, v 2 }, so it has dimension Let A be a matrix with characteristic polynomial (x 1)(x 2)(x + 1)(x + 2). Which need not be true? (a) A is diagonal. (b) There are exactly 24 diagonal matrices with the same characteristic polynomial. (c) The eigenvectors of A span R 4. (d) det(a) = 4. Solution: (a). First note that A must be a 4 4 matrix. (a): The matrix is similar to a diagonal matrix, but may not be diagonal. (b): If a diagonal matrix has the given polynomial as its characteristic polynomial, it must be a 4 4 matrix with diagonal entries 1, 2, 1, 2. There are 24 such matrices. (c): Since the eigenvalues of A are all distinct, the eigenvectors must span R 4. (d): The determinant of A will be the product of its eigenvalues. 10