MIT18_06S10_pset5_s10_soln

# MIT18_06S10_pset5_s10_soln - 18.06 Problem Set 5 Solution...

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Unformatted text preview: 18.06 Problem Set 5 Solution Total: points Section 4.1. Problem 7. Every system with no solution is like the one in problem 6. There are numbers y 1 ,...,y m that multiply the m equations so they add up to = 1. This is called Fredholm’s Alternative: Exactly one of these problems has a solution: A x = b OR A T y = with y T b = 1. If b is not in the column space of A it is not orthogonal to the nullspace of A T . Multiply the equations x 1 − x 2 = 1 and x 2 − x 3 = 1 and x 1 − x 3 = 1 by numbers y 1 ,y 2 ,y 3 chosen so that the equations add up to = 1. Solution (4 points) Let y 1 = 1, y 2 = 1 and y 3 = − 1. Then the left-hand side of the sum of the equations is ( x 1 − x 2 ) + ( x 2 − x 3 ) − ( x 1 − x 3 ) = x 1 − x 2 + x 2 − x 3 + x 3 − x 1 = 0 and the right-hand side is 1 + 1 − 1 = 1 . Problem 9. If A T A x = then A x = 0. Reason: A x is inthe nullspace of A T and also in the of A and those spaces are . Conclusion: A T A has the same nullspace as A . This key fact is repeated in the next section. Solution (4 points) A x is in the nullspace of A T and also in the column space of A and those spaces are orthogonal . Problem 31. The command N=null(A) will produce a basis for the nullspace of A . Then the command B=null(N’) will produce a basis for the of A . Solution (12 points) The matrix N will have as its columns a basis for the nullspace of A . Thus if a vector is in the nullspace of N T it must have dot product with every vector in the basis of N ( A ), thus it must be in the row space of A . Thus the command null(N’) will produce a basis for the row space of A . 1 >> A = rand(6,12) A = Columns 1 through 6 0.8147 0.2785 0.9572 0.7922 0.6787 0.7060 0.9058 0.5469 0.4854 0.9595 0.7577 0.0318 0.1270 0.9575 0.8003 0.6557 0.7431 0.2769 0.9134 0.9649 0.1419 0.0357 0.3922 0.0462 0.6324 0.1576 0.4218 0.8491 0.6555 0.0971 0.0975 0.9706 0.9157 0.9340 0.1712 0.8235 Columns 6 through 12 0.6948 0.7655 0.7094 0.1190 0.7513 0.5472 0.3171 0.7952 0.7547 0.4984 0.2551 0.1386 0.9502 0.1869 0.2760 0.9597 0.5060 0.1493 0.0344 0.4898 0.6797 0.3404 0.6991 0.2575 0.4387 0.4456 0.6551 0.5853 0.8909 0.8407 0.3816 0.6463 0.1626 0.2238 0.9593 0.2543 >> N = null(A); >> B = null(N’) B = 0.1754 0.3288 0.6264-0.0997-0.0309 0.0112-0.1000 0.6512-0.1687 0.5528 0.1346-0.1632 0.4380 0.2124-0.3137-0.1684 0.0050 0.0507-0.0661 0.0094-0.2313-0.5029 0.5792-0.4208 0.4425-0.0527 0.2166 0.0525 0.1071-0.2063 0.2213 0.3024-0.3429-0.2123-0.1149 0.3415 0.6724-0.1323-0.1677 0.1315-0.0249 0.0211-0.0371 0.4083 0.1074-0.3610 0.0993-0.1251 0.1827 0.1595 0.4375-0.0581 0.0910 0.0005 0.1438-0.1870 0.0025 0.4256 0.4134-0.3188-0.0887-0....
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MIT18_06S10_pset5_s10_soln - 18.06 Problem Set 5 Solution...

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