235 Math 1. Short Answer Problems
Term Test 2 Solutions
[2] a) Let B = {v1 , . . . , vk } be an orthonormal basis for a subspace S of an inner product space V . Dene projS and perpS . Solution: Let v V , then projS (v ) and perpS (v ) are the unique vectors such that v = projS (v ) + perpS (v ) where projS (v ) S and perpS (v ) S . In particular, projS (v ) =< v, v1 > v1 + + < v, vk > vk and perpS = v projS (v ). [2] b) Consider the quadratic form Q(x) = 4x2 5x2 + 2x2 3x1 x2 + 4x1 x3 + 2x2 x3 . 1 2 3 Write down the symmetric matrix A such that Q(x) = xT Ax. 4 3/2 2 Solution: A = 3/2 5 1 2 1 2 10 00 , 01 1 1 01 00
[2] c) Find a basis for the orthogonal compliment of the subspace S = Span of M (2, 2) under the standard inner product.
Solution: Observe that dim S = dim M (2, 2) dim S = 4 2 = 2 and that v1 = and v2 = 1 0 are orthogonal to the basis vectors of S . Hence, 11 S = Span{v1 , v2 }.
[2] d) Let Q(x1 , x2 ) = ax2 + 2bx1 x2 + cx2 where ac b2 < 0. Prove that Q is indenite. 1 2 Solution: Let A = ab have eigenvalues 1 and 2 . Then we have 1 2 = det A = bc ac b2 < 0 and so one eigenvalue is positive and the other is negative. Hence Q is indenite.
[2] e) Use the Triangularization theorem to prove that every symmetric matrix is orthogonally diagonalizable. Solution: Let A be a symmetric matrix. Then A has all real eigenvalues, so by the triangularization theorem, we have that there exists an orthogonal matrix P such that P T AP = T where T is upper triangular. But, then T T = (P T AP )T = P T AT P = P T AP = T. Hence, T is symmetric and T must be diagonal. Hence, P orthogonally diagonalizes A.
1
2
222 [6] 2. Let A = 2 2 2. Find an orthogonal matrix P that diagonalizes A and 222 the corresponding diagonal matrix. Solution: Clearly the eigenvalues are 0, 0 and 6. For = 0 we get 22 111 2 2 0 0 0 . 22 000 1 0 . Applying the Gram-Schmidt and w2 = 1 1 1 1 1 1 and v2 = 1. procedure, we get orthonormal eigenvectors v1 = 2 6 0 2 4 2 2 1 0 1 1 1 2 4 2 0 1 1. So, v3 = 1. For = 6 we get A 6I = 3 2 2 4 00 0 1 1/ 2 1/6 1/3 000 Thus, we take P = 1/ 2 1/ 6 1/3 and get D = 0 0 0. 006 0 2 / 6 1/ 3 2 A 0I = 2 2 1 1 Hence, we get eigenvectors w1 = 0
3
[4] 3. Find a and b to obtain the best tting equation of the form y = a + bx for the following data: x 1 0 1 2 y 4 3 1 2 4 3 1 111 Solution: Let AT = and y = . Then we get 1 1 0 1 2 2 a = (AT A)1 AT y 42 26
1
=
4 1 1 1 1 3 1 0 1 2 1 2 6 5/2 = 7 2
= Thus, the line of best t is y =
1 6 2 20 2 4
5 2
2x.
[4] 4. Find an orthonormal basis for P2 using the inner product p(x), q (x) = p(0)q (0) + p(1)q (1) + p(2)q (2), by applying the Gram-Schmidt procedure to the basis {1, x, x2 }. Solution: Let w1 = 1, w2 = x and w3 = x2 . Let v1 = w1 . v 2 = w2 w2 , v1 3 v1 = x (1) = x 1 2 v1 3 w3 , v1 w3 , v2 v 3 = w3 v1 v2 2 v1 v2 2 5 4 1 = x2 (1) (x 1) = x2 2x + 3 2 3 = 2/3 we get an orthonormal basis is
x2 2x+ 1 1 3 , x1 , 3 2 2/3
Then, since x2 2x +
1 3
.
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1 3 1 [7] 5. Let A = 0 1 2 . Find an orthogonal matrix P and upper triangular matrix T 021 T such that P AP = T . 1 0. Thus, Solution: Observe that 1 is an eigenvalue of A with unit eigenvector e1 = 0 3 we can extend {e1 } to the standard basis for R and so we take P = I . Thus P T AP = 12 1 bT where A1 = . 21 0 A1 We have C () = det(A1 I ) = 2 2 3 = ( 3)( + 1). For = 3 we get A1 3I = 2 2 1 1 so v1 = 2 2 00
1 2
1 . 1
1 2
We to need extend {v1 } to an orthonormal basis for R2 , so we take v2 =
1 . 1
1 0 0 2 Thus, we take P1 = 0 1/2 1/ and we get that 0 1/ 2 1/ 2 0 0 1 0 0 1 3 1 1 1 1 2 T 2 2 P1 AP1 = 0 1/2 1/ 0 1 2 0 1/2 1/ = 0 3 0 . 021 0 0 1 0 1/ 2 1/ 2 0 1/ 2 1/ 2
5
6. Let A =
53 and Q(x) = xT Ax. 3 3
[4] a) Find an orthogonal matrix P and diagonal matrix D such that P T AP = D. Solution: We have C () = 5 3 = 2 2 24 = ( 6)( + 4). 3 3 1 3 1 3 , so v1 = 3 9 00 3 . 1
Thus, the eigenvalues of A are 6 and 4. For = 6 we get A 6I = For = 4 we get A + 4I =
1 10
93 31 1 , so v2 = 1 . 10 3 31 00 3/10 1/ 10 60 Hence, we can take P = to get D = . 0 4 1/ 10 3/ 10
[1] b) Classify Q(x) as positive denite, negative denite or indenite. Solution: Since A has a positive eigenvalues and a negative eigenvalue we have that Q(x) is indenite.
[2] c) Write Q so that it has no cross terms and give the change of variables which brings it into this form. Solution: Using the change of variables 3/10 1/ 10 x = Py = 1/ 10 3/ 10
2 2 we get Q(x) = 6y1 4y2 .
y1 (3/10)y1 + (1/10)y2 , = y2 (1/ 10)y1 (3/ 10)y2
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[3] 7. Let A be a positive denite symmetric matrix. Show that A = BB T for some matrix B with orthogonal columns. Solution: Let 1 , . . . , n be the eigenvalues of A which are all positive since A is positive denite. Let D be the diagonal matrix whose diagonal entries are the square roots of the eigenvalues of A. Since A is symmetric, there exists an orthogonal matrix P such that P T AP = D2 = DDT A = P DDT P T = P D(P D)T . Thus, we let B = P D. Hence, the columns of B are just scalar multiples of columns of P and hence the columns of B are orthogonal since the columns of P are orthogonal.
[3] 8. Let A be an m n matrix with n > m and let x Rn . Prove that if AAT is invertible, then the point u Null(A) which is nearest to x is u = x AT (AAT )1 Ax. Solution: By the approximation theorem, the vector u Null(A) which is closest to x is projNull(A) (x). Let v = perpNull(A) (x) so that x = u + v . Then, v Null(A) = Col(AT ). Hence, v = AT t for some vector t Rm . In particular, t is the vector in Col(AT ) which is nearest to x. Hence, by the method of least squares, we have that (AT )T (AT )t = (AT )T x AAT t = Ax, so, t = (AAT )1 Ax. Thus, we have u = x v = x AT (AAT )1 Ax as required.
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[3] 9. Let A be an n n positive denite symmetric matrix and C be an invertible n n matrix. Prove that B = C T AC is also a positive denite symmetric matrix. Solution: Since A is symmetric, B T = (C T AC )T = C T AT C T T = C T AC = B Therefore, B is symmetric. Since A, is positive denite, xT Bx = xT C T ACx = (Cx)T A(Cx) 0 for all x Rn , and xT Bx = (Cx)T A(Cx) = 0 implies that Cx = 0, and hence x = 0 since C is invertible. Thus, B is positive denite.
[3] 10. Let A be a symmetric matrix and let L : Rn Rn be the linear map given by L(x) = Ax. Show that there exists an orthogonal basis B = {v1 , . . . , vn } for Rn such that L is a linear combination of the projections P1 , . . . , Pn where Pi : Rn Rn is the projection Pi (x) = projvi (x) for 1 i n. Solution: Since A is symmetric there exists an orthogonal matrix P = v1 vn and diagonal matrix D such that A = P DP T where the vi are eigenvectors of A corresponding to eigenvalues i for 1 i n. Hence, L(x) = Ax = P DP T x = (P D)(P T x) = 1 v 1 n v n v1 x . . .
vn x = 1 (v1 x)v1 + + n (vn x)vn = 1 projv1 (x) + + n projvn (x)