Differential Equations Solutions 17

Differential Equations Solutions 17 - 27 CHALLENGE 5.12....

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Unformatted text preview: 27 CHALLENGE 5.12. Define c = Q∗ b = c1 c2 R1 0 ,R = , where c1 is n × 1, c2 is (m − n) × 1, R1 is n × n, and 0 is (m − n) × n. Then b − Ax 2 = Q∗ (b − Ax) = c − Rx 2 = c1 − R 1 x = c1 − R 1 x 2 2 2 + c2 − 0x 2 2 + c2 . To minimize this quantity, we make the first term zero by taking x to be the solution to the n × n linear system R1 x = c1 , so we see that the minimum value of b − Ax is c2 . Note that this derivation is based on the three fundamental facts proved in the previous challenge. CHALLENGE 5.13. The two zeros of the function y = norm([.5 .4 a; a .3 .4; .3 .3 .3]) - 1 define the endpoints of the interval. Plotting tells us that one is between 0 and 1 and the other is between 0 and -1. MATLAB’s fzero can be used to find both roots. The roots are −0.5398 and 0.2389. CHALLENGE 5.14. Suppose that {u1 , . . . , uk } form a basis for S . Then any vector in S can be expressed as α1 u1 + . . . + αk uk . Since Aui = λui , we see that A(α1 u1 + . . . + αk uk ) = λ1 α1 u1 + . . . + λk αk uk is also in S , since it is a linear combination of the basis vectors. Therefore, if a subset of the eigenvectors of A form a basis for S , then S is an invariant subspace. Now suppose that S is an invariant subspace for A, so for any x ∈ S , the vector Ax is also in S . Suppose the dimension of S is k , and that some vector x ∈ S has components of eigenvectors corresponding to more than k eigenvalues: r αj uj , x= j =1 where r > k and αj = 0 for j = 1, . . . , r. Consider the vectors x, Ax, . . . , Ar−1 x, all of which are in S , since each is formed from taking A times the previous one. Then x Ax A2 x ... Ar−1 x = u1 u2 u3 ... ur DW, ...
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