Solutions to Second In-Class Exam: Math 401 Section 0201, Professor Levermore Friday, 16 April 2010

Transcription

1 Solutions to Second In-Class Exam: Math 4 Section, Professor Levermore Friday, 6 April No notes, books, or electrontics You must show your reasoning for full credit Good luck! [] Consider the polynomials p (x) = x, p (x) = x( x), p 3 (x) = x( + x) (a) Show that {p (x), p (x), p 3 (x)} is a basis for P () (b) Express q(x) = + 3x in terms of this basis Solution (a) The set {p (x), p (x), p 3 (x)} is related to the standard basis {, x, x } of P () by ( p (x) p (x) p 3 (x) ) = ( x x ) Then {p (x), p (x), p 3 (x)} will be a basis for P () if and only if the matrix above is invertible There are many ways we can show this For example, we can show that det = ( ) =, or we can use row reduction to show that The invertiblity of the matrix then implies that {p (x), p (x), p 3 (x)} is a basis for P () Alternative Solution (a) It is clear that {p (x), p (x), p 3 (x)} P () Because the dimension of P () is three, we only have to show that the set {p (x), p (x), p 3 (x)} is linearly independent We do this by showing that c p (x) + c p (x) + c 3 p 3 (x) = = c = c = c 3 = This can be done several ways For example, upon evaluating the linear combination at x =, x =, and x = we obtain = c p () + c p () + c 3 p 3 () = c, = c p ( ) + c p ( ) + c 3 p 3 ( ) = c, = c p () + c p () + c 3 p 3 () = c 3, which implies that c = c = c 3 = Alternatively, if we express the linear combination in terms of the standard basis {, x, x } of P () it becomes = c ( x ) + c (x x ) + c 3 (x + x ) = c + (c + c 3 )x + ( c c + c 3 )x This leads to the system c =, c + c 3 =, c + c c 3 =, which can be solved to show that c = c = c 3 =

2 Solution (b) By setting we are led to the system + 3x = q(x) = c p (x) + c p (x) + c 3 p 3 (x) = c ( x ) + c (x x ) + c 3 (x + x ) = c + (c + c 3 )x + ( c c + c 3 )x, c =, c + c 3 = 3, c + c c 3 =, which can be solved to find that c = c = and c 3 = We thereby obtain [5] Consider the directed graph q(x) = p (x) + p (x) + p 3 (x) ր 4 3 (a) Label the edges and give the corresponding incidence matrix A for the graph (b) Give a basis for the ker(a T ) Solution (a) If we label the five edges a, b, c, d, and e as a d e ր b 4 3 c then the corresponding incidence matrix A is A = Here we used the convention that the vertex at the base of the arrow along an edge is assigned while the vertex at the point is assigned Had we flipped this convention then A would be the negative of the matrix given above Solution (b) Because A T is 5 4 while rank(a T ) =rank(a) = 4 = 3, we know that ker(a T ) is two dimensional By row reduction we see that A T =

3 We see there are two free parameters in the general solution, which is what we expect because ker(a T ) is two dimensional We can obtain a basis for ker(a T ) by setting each parameter to while setting the other to This gives the basis, The first of these corresponds to the cycle abcd ( 3 4 ), while the second corresponds to the cycle bce ( 3 4 ) a b 3 [] Define the linear mapping L(X) = AX for every matrix X where A = b a A basis for matrices is given by E =, E =, E 3 =, E 4 = Give the matrix representative of L with respect to this basis Solution Given any finite dimensional linear space U and linear mapping L : U U, the matrix representative of L with respect to a basis {u,u,,u m } of U is the m m matrix R such that ( L(u ) L(u ) L(u m ) ) = ( u u u m ) R For the given basis {E, E, E 3, E 4 } of matrices, direct calculations show that a b a b L(E ) = AE = = = ae b a b a + be 3, a b a b L(E ) = AE = = = ae b a b a + be 4, a b b a L(E 3 ) = AE 3 = = = be b a a b + ae 3, a b b a L(E 4 ) = AE 4 = = = ae b a a b + be 4 Because these calculations show that ( L(E ) L(E ) L(E 3 ) L(E 4 ) ) = ( a b ) E E E 3 E 4 a b b a, b a the matrix representative of L with respect to the basis {E, E, E 3, E 4 } is a b a b b a b a 3

4 4 4 [5] Construct monic polynomials p, p, p, and p 3 of degrees,,, and 3 respectively that are orthogonal with respect to the inner product f, g = f(x) g(x) x 4 dx Solution By applying the Gram-Schmidt procedure to {, x, x, x 3 } you obtain the monic orthogonal polynomials p (x) =, p (x) = x, x,, p (x) = x, x, p, x p, p p (x), p 3 (x) = x 3, x3, p, x 3 p, p p (x) p, x 3 p, p p (x) Because their integrands each have odd symmetry over [, ], you see that while direct calculations show that, x = x, x =, x 3 = x, x 3 =,, =, x = x, x = x, x 3 = x, x = The Gram-Schmidt procedure therefore yields p (x) =, p (x) = x = x, x 4 dx = 5 x5 x 6 dx = 7 x7 x 8 dx = 9 x9 p (x) = x 5 7 x = x 5 7, = 5, = 7, = 9 p 3 (x) = x x (x 5 7 ) = x3 7 9 x, where in the last step we used the fact that p, x 3 = x, x 3 5 7, x3 = 5 [5] Find all real values of c for which the matrix A = c is Hermitian positive Be sure to give your reasoning! Solution Because c is real, the matrix A is Hermitian It will therefore be Hermitian positive if and only if the determinants of its principle minors are all positive These

5 determinants are det() =, det = c, det c = c c It is clear that all of these determinants are positive if and only if c > Therefore A is Hermitian positive if and only if c > Alternative Solution Because c is real, the matrix A is Hermitian It will therefore be Hermitian positive if and only if all of its pivots are positive By row reduction we obtain A = c c c, c where we must assume that c The pivots are given by, c, c = c c It is clear that all of these pivots are positive if and only if c > Therefore A is Hermitian positive if and only if c > Remark The two approaches to the solution given above are related by the general fact that the n th pivot of A for n > is given by det(a n )/ det(a n ), where A n denotes the n th principle minor matrix of A, while the first pivot of A is a = det(a ) So showing that the pivots all are positive is equivalent to showing that the principle minors all have a positive determinant 6 [5] Consider the plane W R 4 spanned by ( )T and ( )T Equip R 4 with the Euclidean inner product (a) Find the point in W that is closest to the point ( ) T (b) Give the shortest distance from W to the point ( ) T Solution (a) Let w =, w =, and u = The point in W closest to u is given by the orthogonal projection of u onto W Because w T w =, w T w = w T w = 4, w T u =, w T u =, we see that {w,w } is an orthogonal basis for W and that the orthogonal projection of u onto W is therefore given by Pu = wt u w Tw w + wt u w Tw w = w + w = w The point w in W that is closest to u is thereby w = Pu = w = ( ) T 5

6 6 Alternative Solution (a) Let A =, and u = The point in W that is closest to u is given in terms of the least least squares solution with respect to the Euclidean inner product of the overdetermined system c A = u c Specifically, the point w in W that is closest to u is given by c w = A, where A c T c A = A c T u Because A T A = 4 =, 4 A T u = = we see that ( 4 c =, 4 c ) whereby c = and c = The point in W that is closest to u is thereby ) w = A ( = ( ) T Solution (b) The shortest distance from W to u is u w where u w = = One thereby sees that this distance is u w = ( ) + ( ) + ( ) + ( ) = = =