quiz2 and ans

quiz2 and ans - invariant under multiplication by Q so k y...

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AMSC/CMSC 660 Quiz 2 , Fall 2006 1. (10) Suppose we have factored the m × n matrix A = QR ( m n ), and let ˆ x be the solution to the least squares problem min x k Ax - b k . Show that k A ˆ x - b k 2 = k c 2 k 2 , where c 2 is the vector consisting of the last m - n components of Q * b . Answer: (Note that we must assume that A is full rank.) De±ne c = Q * b = c 1 c 2 , R = R 1 0 where c 1 is n × 1, c 2 is ( m - n ) × 1, R 1 is n × n , and 0 is ( m - n ) × n . Then k Ax - b k 2 = k Q * ( Ax - b ) k 2 = k Rx - c k 2 = k R 1 x - c 1 k 2 + k 0x - c 2 k 2 = k R 1 x - c 1 k 2 + k c 2 k 2 . To minimize this quantity, we make the ±rst term zero by taking x to be the solution to the n × n linear system R 1 x = c 1 , so we see that the minimum value of k Ax - b k is k c 2 k . Note: The derivation above is based on three fundamental facts: Minimizing the norm of Ax - b gives the same solution as minimizing the square of the norm. For any vector y and any unitary matrix Q the norm of the vector is
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Unformatted text preview: invariant under multiplication by Q , so k y k = k Q * y k . • Suppose we partition the vector y into two pieces: y = • y 1 y 2 ‚ . Then k y k 2 = k y 1 k 2 + k y 2 k 2 . 1 2. (10) Write a column-oriented algorithm to solve Ux = b where U is an n × n nonsingular upper triangular matrix. (If you can’t do this, you can get 5 points for any correct algorithm to solve this problem, but you may not use the backslash operator or an inverse matrix.) Answer: (Note that this is like the 4th exercise.) x = b; for j=n:-1:1, x(j) = x(j) / U(j,j); x(1:j-1) = x(1:j-1) - U(1:j-1,j)*x(j); end 2...
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This homework help was uploaded on 02/05/2008 for the course CMSC 660 taught by Professor Oleary during the Fall '06 term at Maryland.

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quiz2 and ans - invariant under multiplication by Q so k y...

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