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Dec2004Solutions

# Dec2004Solutions - MAT223 Final December 2004 Note I’ll...

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Unformatted text preview: MAT223 Final December 2004 Note: I’ll often write column vectors horziontally and with commas to save space. Part I 1. (2). Solving for X , X = (AT)− 1A − 1 = (A AT) −1. Now, det A = 1, so det(A AT) = 1, which ab d −b means that the inverse of A AT = is just , and we only need to compute cd −c a 1 a=[1 −1] = 2. −1 2. (4). In A we have the solutions of a homogeneous system of linear equations, so there we do have a subspace. In B the system is not linear and though in C it is, it’s not homogeneous. 0 −1 1 3. (1). In the basis {1, x, x2}, the vectors in S are the rows of − 1 1 0 whose determi1 k1 nant is − (k + 2). Thus S is a basis unless k = − 2. (Alternatively, use row-reduction.) 4. (1). The given basis for W is orthogonal, so we can compute X · V2 1 1 1 1 1 X · V1 V1 + V2 = − 1 + 1 = 0 . projW X = V2 2 20 V1 2 20 0 5. (5). X = V3 − V3 · V1 V1 2 V1 + V3 · V2 V2 2 1 2 1 V2 = [1, 1, 0, 1] − 3 [1, − 1, 0, 1] − 2 [1, 1, 0, 0] = 3 [ − 1, 1, 0, 2]. 6. (2). You get the elemntary matrix by applying the same row operation to the identity matrix. 7. (6). By Cramer’s Rule, x2 = a − 2b c d − 2e f g − 2h k / abc def ghk =−2 abc def ghk / abc def ghk = − 2. 8. (5). A is false since the eigenvalues of A− 1 are λ− 1 where λ is an eigenvalue of A, so they are usually diﬀerent from those of A. B is true: if A v = λ v , then A− 1v = λ− 1v , so (A− 1 + 1+λ I )v = λ− 1v + v = λ v . C is false, not every invertible matrix is diagonalizable. 9. (5). Expanding by the second row, det A = − 2k 2 2 − 2k (k − 1) (k + 1) − 1 = 2k (k − 1)(k + 2). 1 k−1 k+1 k + 1 k2 1 0 k−1 0 = 2k (k − 1) 1 k+1 k+1 1 = 10. (4). A is true: if A B is invertible, B − 1 = (A B )− 1A exists. B is also true: if A B = − (A B )T, taking determinants on both sides, det A det B = ( − 1)3 det A det B , and since det A 0, it follows that det B = 0. C is false: if A B = B A, B could be the zero matrix, for example. 11. (6). Since dim(im T ) + dim(ker T ) = dim(domain) = 7, option A is correct. B is not because, the rank of A is dim(im T ), not 8 minus that. Finally, C is also correct because rank A = dim(row A) = 7 − dim((row A)⊤). 12. (1). The third column of the standard matrix representation is just T ([0, 0, 1]) = 2[0, 0, 1] − N = [ − 1, 0, 1]. 1 Part II 13. a) If B is not invertible, det B = 0, that is, a d − b c = 0. Therefore, B2 = a2 + a d a b + b d a2 + b c a b + b d ab = (a + d) = (a + d)B. = 2 cd a c + c d a d + d2 ac+cd bc+d b) Assume that t1 X1 + t2 X2 + t3 X3 = 0, we must prove t1 = t2 = t3 = 0. Well, multiplying by the matrix A we get t1 A X1 + t2 A X2 + t3 A X3 = 0. Since {A X1, A X2, A X3} is assumed to be independent, we get t1 = t2 = t3 = 0 as desired. The converse is false in general, for example if A is the zero matrix, clearly {A X1, A X2, A X3} will de linearly dependent no matter what. 14. a) Note that W ⊥ is given as a solution set of a system of homogeneous linear equations, thus we can write it as the null space of the matrix of coeﬃcients of said 1 0 −1 0 0 1 1 0 0 −1 system: W ⊥ = null A where A = 0 1 1 0 − 1 . So we want a basis for W = 0 0 0 1 −1 ⊥⊥ ⊥ (W ) = (null A) . I claim that (null A)⊥ = row A. Consider an arbitrary X ∈ null A. To calculate the product A X , we take the dot product of each row of A with the column vector X , and since X ∈ null A, all of these products must be zero. This means that each row of A belongs to (null A)⊥, and so, row A ⊆ (null A)⊥. To prove those subspaces are equal it is enough to show they have the same dimension, but dim (null A)⊥ = 5 − dim null A = rank A = dim row A. So, we have W = row A. To ﬁnd a basis for W , we simply use row reduction: 1 1 0 0 0 −1 10 11 00 0 0 0 1 0 1 −1 R2 − R1 0 0 −1 −1 0 0 −1 11 11 00 0 0 0 1 0 1 −1 R3 − R2 0 0 −1 −1 0 0 −1 11 00 00 0 0 0 1 0 −1 , 0 −1 so we see that {[1, 0, − 1, 0, 0], [0, 1, 1, 0, − 1], [0, 0, 0, 1, − 1]} is a basis for W . b) Let X ∈ W ⊤, we must show X ∈ U ⊤. For this we have to show that X · Y = 0 for any Y ∈ U . But any Y ∈ U is also in W (since U ⊆ W ), and since X ∈ W ⊤, we do have X · Y = 0. 15. a) We need eigenvectors corresponding to the eigenvalues 2 and − 1. First, let’s do the eigenvalue − 1: we need to solve (A − ( − 1)I )X = 0, that is, (A + I )X = 0. The three equations in the system are the same, namely, x + y + z = 0. The general solution, of course, is given by x = t1, y = t2, z = − t1 − t2. We ﬁnd a basis of the solution space by setting t1 = 1, t2 = 0 to get [1, 0, − 1] and then t1 = 0, t2 = 1 to get [0, 1, − 1]. No for the eigenvalue 2: we solve the system (A − 2I )X = 0 by row reduction: 1 −2 1 1 − 2 1 R3 + R2 1 − 2 1 −2 1 1 1 − 2 1 R1 ↔ R2 − 2 1 1 R2 + 2R1 0 − 3 3 1 0 1 − 1 R3 − R1 1 1 −2 0 3 − 3 − 3 R2 0 0 0 1 1 −2 2 which means that the solution space is given by z = t, y = t, x = t, so [1, 1, 1] is a basis for it. 1 01 Therefore, writing our eigenvectors as columns, P = 0 1 1 , and we’ll have −1 −1 1 −1 0 0 P − 1A P = 0 − 1 0 . A is diagonalizable because we found a basis of eigenvec0 02 tors: the columns of P . a00 b) Let P be such that P − 1A P is a diagonal matrix, say 0 b 0 . Then we have 00c 3 a3 0 0 a00 0 b3 0 = 0 b 0 = (P − 1A P )3 = P − 1A3P = 0, 00c 0 0 c3 so a3 = b3 = c3 = 0. But this implies that a = b = c = 0, and so P − 1A P = 0. Solving for A, A = P 0 P − 1 = 0. 16. a) Clearly both A and B have two dimensional columns spaces. By either row reduction, or a little trial and error, we ﬁnd that the columns of B are linear combinations of the columns of A: B1 = 2A1 − A2 and B2 = A1 + A2. This tells us that col B ⊆ col A. Since both spaces have dimension two, they must be equal. b) Let X ∈ null A ∩ col A. Since X ∈ col A = im A, we know that X = A Y for some vector Y . We also have X ∈ null A, so that A X = 0, i.e., A2Y = 0, so that Y ∈ null A2 = null A. Therefore, X = A Y = 0, as desired. 3 ...
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