Unformatted text preview: j + u j,j +1 y j +1 + ... + u jn y n = b j , and since at this point we know y j +1 ,...,y n , we can solve this for y j : y j = ( b j-n X i = j +1 u ji y i ) / u jj . The pseudocode now might be for j = n : -1 : 1 do the following: y(j) = ( b(j) - u(j,j+1:n) * y(j+1:n) ) / u(j,j) end for j That’s it! It is that simple: y(j) = (scalar - dot product ) / scalar Of course this only works when the diagonal of U has no zero elements; but that is precisely when U is nonsingular. Therefore (in exact arithmetic, at least), this method always returns the solution when a unique solution exists. What about ﬁnite precision arithmetic? It does remarkably well, in fact if ¯ y is the computed solution, then there exists a matrix E such that ( U + E )¯ y = b, where | E | ≤ n μ | U | + O( μ 2 )...
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- Fall '10
- Diagonal matrix, Triangular matrix, Square Triangular Systems, form unn yn