UNIVERSITY OF TORONTO AT SCARBOROUGH DEPARTMENT OF COMPUTER AND MATHEMATICAL SCIENCES Final Examination Solutions MATB24H3 Linear Algebra II Examiner: L. Zhao Date: Monday 20 December 2004 Duration: Three(3) Hours, 14:00 - 17:00 (1) Let V be a real vector space. (a) (3 Points) Give the definition for a subset W of V to be a subspace of V . A subset W of V is a subspace of V if W is also a real vector space, where addition and scalar multiplication of vectors in W produce the same vectors as these operations do in V . (b) (3 Points) Give the definition of an inner product of V . An inner product on V is a function that assigns each ordered pair of vectors ~v and w~ in V a real number, written < ~v; w~ >, satisfying the following properties, for all ~v; w~; ~u V and r R: (i) <2~v; w~ >=<2w~; ~v >. (ii) < ~v; w~ + ~u >=< ~v; w~ > + < ~v; ~u >. (iii) r < ~v; w~ >=< r~v; w~ >=< ~v; rw~ >. (iv) < ~v; ~v > 0 and < ~v; ~v >= 0 if and only if ~v = ~0. (c) (3 Points) Assuming≥ that an inner product has been defined for V , give the definition of an orthonormal basis of V . An orthonormal basis of V is a basis B of V such that < ~b;~b >= 1 for all ~b B and 2 < ~b; b~ >= 0 for all ~b; b~ B and ~b = b~ . 0 0 2 6 0 (d) (3 Points) Using the definition of an inner product, verify that for ~v1 = (v1; v2) and 2 2 ~u = (u1; u2) in R , < ~v; ~u >= 3v1u1 + 2v2u2 is an inner product on R . Let ~v; w~; ~u R2 and r R. First, we have 2 2 < ~v; ~u >= 3v1u1 + 2v2u2 = 3u1v1 + 2u2v2 =< ~u; ~v > : Second, we have < ~v; w~ +~u >= 3v1(w1 +u1)+2v2(w2 +u2) = 3v1w1 +2v2w2 +3v1u1 +2v2u2 =< ~v; w~ > + < ~v; ~u > : Third, we have < r~v; ~u >= 3rv1u1 + 2rv2u2 = r < ~v; ~u >; and < ~v; r~u >= 3v1ru1 + 2v2ru2 = r < ~v; ~u > : Fourth, we have < ~v; ~v >= 3v2 + 2v2 0; 1 2 ≥ since squares are always non-negative and the equality hold in the above expression if and only if v1 = v2 = 0 or ~v = ~0. Hence the formula we have does give rise to an inner product. (e) (3 Points) Find vectors ~v1 = (v1; v2) and ~u = (u1; u2) that are orthonormal with re- 2 spect of the inner product R , < ~v; ~u >= 3v1u1 + 2v2u2 but are not orthonormal with respect to the dot product on R2. 2 MATB24H3 LINEAR ALGEBRA II Let ~v = (1=p3; 0) and ~u = (0; 1=p2). It is obvious that these vectors are not orthonor- mal since either has normal zero in the dot products. However, it can be easily checked that 1 1 < ~v; ~u >= 3 0 + 2 0 = 0; × p3 × × × p2 while 1 1 < ~v; ~v >= 3 + 2 0 0 = 1; × p3 × p3 × × and 1 1 < ~u; ~u >= 3 0 0 + 2 = 1: × × × p2 × p2 These are certainly not the only examples. (2) Let W be the subspace of R3 generated by the vectors [1; 1; 1] and [2; 1; 1]. 1 2 (a) (5 Points) Find the projection matrix for projection into W . Let A = 1 1 . The 0 1 1 1 T 1 T projection matrix we are looking for is P = A(A A)− A . First @ A 1 1 2 − 1 T 1 1 1 1 3 4 − 1 6 4 (A A)− = 1 1 = = − : 2 2 1 1 0 13 4 6 2 4 3 1 1 − Hence 4 @ A5 2 0 0 T 1 T 1 1 1 1 6 4 1 1 1 1 P = A(A A)− A = − = 0 1 1 : 2 2 1 1 4 3 2 1 1 2 0 1 − 0 1 1 (b) (3 Points) Find the projection of the vector ~v = [ 2; 1; 2] on W@. A − − The project of ~v on W can easily be computed as 2 0 0 2 4 1 1 P~v = 0 1 1 −1 = −1 : 2 0 0 1 1 1 0 −2 1 2 0 1 1 @ A @ A @ A (c) (2 Points) Find ~bW and ~bW ? such that ~bW W , ~bW ? W ? and ~v = ~bW +~bW ? , where ~v is as in part (b). 2 2 The vector computed in part (b) is ~bW and 1 3 3 ~bW ? = ~v ~bW = [ 2; 1; 2] [ 4; 1; 1] = 0; ; : − − − − 2 − −2 2 (3) (10 Points) Transform the basis [1; 1; 0]; [2; 1; 3]; [2; 2; 2] for R3 into an orthonormal basis for R3 using the Gram-Schmidt prof cess. g Let ~a1 = [1; 1; 0], ~a2 = [2; 1; 3], and ~a3 = [2; 2; 2]. First, let ~v1 = ~a1 = [1; 1; 0]. Next, let ~a ~v 3 1 1 ~v = ~a 2 · 1 ~v = [2; 1; 3] [1; 1; 0] = ; ; 3 : 2 2 − ~v ~v 1 − 2 2 −2 1 · 1 Last, let ~a ~v ~a ~v 4 6 1 1 1 ~v = ~a 3 · 1 ~v 3 · 1 ~v = [2; 2; 2] [1; 1; 0] ; ; 3 = [6; 6; 2]: 3 3 − ~v ~v 1 − ~v ~v 1 − 2 − 19=2 2 −2 −19 − 1 · 1 1 · 1 FINAL EXAMINATION SOLUTIONS 3 The vectors ~vi with 1 i 3 for an orthogonal basis of R3. To find an orthonormal basis, it remains to normalized≤ all≤ the vectors, we have the follow three vectors is an orthonormal basis of R3: 1 1 1 [1; 1; 0]; [1; 1; 6]; [3; 3; 1]: p2 p38 − p19 − x t (4) (15 Points) Find the matrix C such that the = C 1 would diagonalize the y t2 following quadratic form. Also, find the diagonalized form ofthe quadratic form. f(x; y) = 10x2 + 6xy + 2y2: Sketch the graph of the conic section f(x; y) = 4. 10 3 We write down the symmetric coefficient matrix A = . To find the eigenvalues, 3 2 we write down the characteristic equation of A. We have det(A λI) = (10 λ)(2 λ) 9 = λ2 12λ + 11 = (λ 1)(λ 11): − − − − − − − Hence the eigenvalues of A are λ1 = 1 and λ2 = 11. For λ1 = 1, 9 3 A I = : − 3 1 1 1 1 Hence an eigenvector for λ1 = 1 is − or − upon normalizing. 3 p10 3 For λ2 = 11, 1 3 A 11I = − : − 3 9 − 3 3 Hence an eigenvector for λ = 11 is or 1 upon normalizing. 2 1 p10 1 − − Therefore, the change of variables matrix C that diagonalizes the given quadratic form is 1 1 3 C = − ; p 3 1 10 − and the diagonalized quadratic form is 2 2 x 1 1 3 t1 t1 + 11t2; with = − : y p10 3 1 t2 − The conic section we are to sketch can be re-written as 10x2 + 6xy + 2y2 4 = 0. With the change of variables give above, it becomes − t2 t2 t2 + 11t2 4 = 0; or equivalently 1 + 2 = 1: 1 2 − 22 (2=p11)2 This is an ellipse, centered at the origin that crosses the t1-axis at 2 and the t2-axis at 2=p11. In the x y plane the t -axis goes through the origin and the point ( 1; 3) with − 1 − its positive direction pointing into the second quadrant; the t2-axis goes through the origin and the point ( 3; 1) with its positive direction pointing into the third quadrant. − − 1 2 i 0 (5) (10 Points) Let A = 2 + i −1 0 . Find a unitary matrix U and diagonal matrix 0 0 −0 3 1 1 D such that D = U − @AU. A 4 MATB24H3 LINEAR ALGEBRA II We first compute all the eigenvalues of A. We have det(A λI) = [(1 λ)( 1 λ) 5](3 λ) = (λ2 1 5)(3 λ) = (λ2 6)(3 λ): − − − − − − − − − − − Therefore, the eigenvalues are λ = 3 and p6. Hence all the eigenvalues are distinct. We also note that A is Hermitian from which we can infer that the eigenvectors associated with different eigenvalues are orthogonal. Therefore, the normalized eigenvectors of A form an orthonormal basis of C3 with which we can use to form the columns of U. Now it remains to compute and normalize the eigenvectors of A. For λ = p6, we have 1 p6 2 i 0 − − A p6I = 2 + i 1 p6 0 : − 0 − − 1 0 0 3 p6 − @ A An eigenvector for λ = p6 is 5 1 5 2 − 1 p6 p6 1 − 1 p6 2 +− i ; or − 2 +− i ; 0 1 10p6 0 1 0 ! 0 @ A @ A upon normalizing. For λ = p6, we have − 1 + p6 2 i 0 − A + p6I = 2 + i 1 + p6 0 : 0 − 1 0 0 3 + p6 @ A An eigenvector for λ = p6 is − 5 1 5 2 − 1+p6 p6 + 1 − 1+p6 2 + i ; or 2 + i ; 0 1 10p6 0 1 0 ! 0 upon normalizing. @For λ = 3, Awe have @ A 1 3 2 i 0 A 3I = 2 −+ i 1− 3 0 : − 0 0 − 0− 0 1 An eigenvector for λ = 3 is @ A 0 0 ; 0 1 1 and this vector is already of norm one.@ Therefore,A we have p6 0 0 D = 0 p6 0 ; 0 1 0 −0 3 and @ A 5 5 − 0 p10(6 p6) p10(6+p6) 1− 1 U = 0 p6 1 2 p6+1 2 1 : − (2 + i) (2 + i) 0 10p6 10p6 B 0 0 1 C B C @ 3 0 1 A (6) (10 Points) Let A = −2 2 1 . Find a Jordan canonical form J and a matrix C 0 1 −0 1 1 1 − − such that J = C− AC@.