CAAM 335: Matrix Analysis

Transcription

1 1 CAAM 335: Matrix Analysis Solutions to Problem Set 4, September 22, 2008 Due: Monday September 29, 2008 Problem 1 (10+10=20 points) (1) Let M be a subspace of R n. Prove that the complement of M in R n, defined as M = {x R n : x T y = 0 for all y M}, is also a subspace of R n. (2) Let M R 4 be spanned by {(1,2,2,3) T,(1,3,3,2) T,(0,1,1, 1) T }. Find a basis for M. (1) We must show that the vectors in M satisfy the two properties of a subspace. Consider two vectors, p,q M, then (p + q) T y = p T y + q T y = = 0 for all y M, where we have distributed and used the fact that both p and q are in M. Now consider p M and any scalar α R, (αp) T y = α(p T y) = α 0 = 0 for all y M, where we have used the fact that p M. Both properties are satisfied, thus M is a subspace. (2)Define S = We have R (S) = M. By the Fundamental Theorem of Linear Albegra, M = R (S) = N (S T ). S T = = (S T ) red , 1 0 is a basis for N (S T ). 0 1

2 2 Problem 2 ( =25 points) We wish to show that N (A) = N (A T A) regardless of A. 1. We first take a concrete example. Report the findings of the MATLAB null command applied to A and A T A for the matrix A associated with Figure 3.1 in the class notes. 2. For arbitrary A show that N (A) N (A T A), i.e., that if Ax = 0 then A T Ax = For arbitrary A show that N (A T A) N (A), i.e., that if A T Ax = 0 then Ax = 0. (Hint: if A T Ax = 0, then x T A T Ax = 0 and says something about Ax 2, where y 2 2 = yt y.) 4. Let K R m m be a diagonal matrix with positive diagonal entries and let A R m n. Show that N (A) = N (A T KA). (Hint: A T KA = à T Ã. What is Ã?) 1. The output of null shows that A has a two dimensional nullspace corresponding to vertically displacing nodes 1&3 and nodes 1&4. disp(null(a)) If Ax = 0 then A T Ax = A T 0 = 0. Hence N (A) N (A T A). 3. If A T Ax = 0 then x T A T Ax = (Ax) T (Ax) = 0. For any vector y, y T y = 0 implies y = 0. Hence x T A T Ax = (Ax) T (Ax) = 0 implies Ax = 0. Thus, N (A T A) N (A). 4. Let K = diag(k 1,...,k m ) and define K 1/2 = diag( k 1,..., k m ). Then K = K 1/2 K 1/2. Furthermore A T KA = A T K 1/2 K 1/2 A = ((K 1/2 ) T A) T (K 1/2 A) = ((K 1/2 )A) T (K 1/2 A) since K 1/2 is symmetric. If we define à = K 1/2 A. Then A T KA = à T Ã. By parts (b,c), we have N (à T Ã) = N (Ã). Moreover, since k 1,...,k m > 0, we have Ãx = K 1/2 Ax = 0 if and only if Ax = 0. Hence N (A T KA) = N (à T Ã) = N (Ã) = N (A).

3 3 Problem 3 ( =40 points) Write down a matrix with the required properties or explain why no such matrix exists. 1. The column space contains (1,0,0) T,(0,0,1) T, while the row space contains (1,1) T,(1,2) T. 2. The column space has basis {(1,1,1) T }, while the row space has basis {(1,2,1) T }. 3. The column space is R 4 while the row space is R The column space is perpendicular to the null space. 1. If there exists such a matrix A it must be in R 3 2. Moreover, since (1,0,0) T,(0,0,1) T are linearly independent, such a matrix must satisfy dim(r (A)) = 2 = dim(r (A T )) and, hence R (A T ) = R 2. Consider the matrix A = {(1,0,0) T,(0,0,1) T } is a basis for R (A). Moreover, R (A T ) = R 2. Hence A has the desired properties. 2. The column space has basis {(1,1,1) T }, while the row space has basis {(1,2,1) T }. Consider the matrix A = = 1 1 (1,2,1) {(1,1,1) T } is a basis for R (A) and {(1,2,1) T } is a basis for R (A T ). 3. Since dim(r (A)) = r and dim(r (A T )) = r there does not exists a matrix with R (A) = R 4 and R (A T ) = R By the Fundamental Theorem R n = R (A T ) N (A), R (A T ) N (A). Dimension of R (A T ) = r. Dimension of N (A) = n r. R m = R (A) N (A T ), R (A) N (A T ). Dimension of R (A) = r. Dimension of N (A T ) = m r. If A is symmetric, i.e., A = A T then R n = R (A) N (A), R (A) N (A). Dimension of R (A) = r. Dimension of N (A) = n r. Hence any symmetric matrix has the desired property. In particular for the ( ) 1 0 A = {(1,0) T } is a basis for R (A), {(0,1) T } is a basis for N (A) and R (A) N (A).

4 4 Problem 4 (5+10=15 points) Let A R be the matrix corresponding to the truss (the tissue model) in Figure 2.3 of the notes. The matrix A is generated by the MATLAB program fiber.m posted on the CAAM335 References page under Lecture 7. Let K = diag(k 1,...,k 20 ) R be a diagonal matrix with positive diagonal entries k 1,...,k 20 > Use the MATLAB command null to compute a basis for N (A). 2. Use N (A) = N (A T KA) and the Fundamental Theorem of Linear Algebra to decide for which of the two right hand sides specified below the linear system (A T KA)x = f has a solution. (You can t compute the solution, since you do not know K.) f = [-1;1;0;1;1;1;-1;0;0;0;1;0;-1;-1;0;-1;1;-1] f = [1;0;1;0;1;0;1;0;1;0;1;0;1;0;1;0;1;0] (Note: Even if two vectors x,y satisfy x T y = 0 in exact arithmetic, MATLAB x *y may produce a nonzero number. Use abs(x *y) < 1.e-12 to decide whether x T y = 0.) By the Fundamental Theorem of Linear Algebra, R (A T KA) N (A T KA) and R n = R (A T KA) N (A T KA). Hence (A T KA)x = f has a solution if and only if f N (A T KA), i.e., if and only if f T y for all vectors y in a basis for N (A T KA). Since N (A) = N (A T KA), we don t need to compute A T KA, but can compute a basis of N (A). The matlab code follows: % Generate A fiber % Compute a basis for N(A). Store the basis % vectors as columns of B. B = null(a); fprintf( A basis for N(A) \n ) disp(b) f = [-1;1;0;1;1;1;-1;0;0;0;1;0;-1;-1;0;-1;1;-1]; fprintf( B *f = \n ) fprintf( %12.6e \n, B *f) if( any(abs(b *f) > 1.e-12) ) fprintf( The right hand side is not orthogonal to N(A *K*A) \n ) else fprintf( The right hand side is orthogonal to N(A *K*A) \n )

5 5 end f = [1;0;1;0;1;0;1;0;1;0;1;0;1;0;1;0;1;0]; fprintf( B *f = \n ) fprintf( %12.6e \n, B *f) if( any(abs(b *f) > 1.e-12) ) fprintf( The right hand side is not orthogonal to N(A *K*A) \n ) else fprintf( The right hand side is orthogonal to N(A *K*A) \n ) end Note, the computations below were done using Matlab Version (R2008a). The null command on older Matlab versions computes a different basis for the null-space of A. prob4_hw4 A basis for N(A) B *f = e e e-17 The right hand side is orthogonal to N(A *K*A) B *f = e e e+00 The right hand side is not orthogonal to N(A *K*A) The linear system (A T KA)x = f with the first right hand side has a solution, the linear systems with the second right hand side does not have a soltion.