If we count up the operations above we have to eliminate N 1 N 2 terms each at

# If we count up the operations above we have to

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If we count up the operations above, we have to eliminate ( N - 1) N/ 2 terms, each at a cost of O ( N ), so computing the LU factorization is O ( N 3 ) (same as using Gaussian elimination to solve Ax = b 0 for a particular b 0 ). 20 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019 Subscribe to view the full document.

Cholesky factorization The basic idea in Cholesky factorization is that you can use elimi- nation to find an lower-triangular matrix R 1 that eliminates all but the first entry in the first column of A : R 1 a 11 a 12 · · · a 1 N a 21 a 22 · · · a 2 N . . . . . . a N 1 a N 2 · · · a NN = a 11 a 0 12 · · · a 0 1 N 0 a 0 22 · · · a 0 2 N . . . . . . 0 a 0 N 2 · · · a 0 NN . Since A is symmetric, we can do the same elimination on the columns to get R 1 AR T 1 = 1 0 · · · 0 0 a 00 22 · · · a 00 2 N . . . . . . 0 a 00 N 2 · · · a 00 NN = 1 0 T 0 A 1 It is easy to see that A 1 must also be symmetric (and positive defi- nite), so we continue by eliminating all but the first entry in its first column using another lower-triangular matrix R 2 , then doing the same to the columns to get R 2 R 1 AR T 1 R T 2 = I 0 0 A 2 . After N such iterations, we have R N · · · R 1 AR T 1 · · · R T N = RAR T = I . Since the product of two lower-triangular matrices is again lower triangular (show this on your own!), R is again lower-triangular. Since the inverse of a lower-triangular matrix is again lower triangular (show this on your own!), we can take A = LL T , with L = R - 1 . 21 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019 Symmetric QR factorization We describe one algorithm for computing a symmetric QR factor- ization which is very efficient and fairly easy to understand, called Household tri-diagonalization . Let A be a symmetric matrix A = a 11 a 12 · · · a 1 N a 21 a 22 · · · a 2 N . . . . . . . . . a N 1 a N 2 · · · a NN , and set x 1 = 0 a 21 a 31 . . . a N 1 , y 1 = 0 r 1 0 . . . 0 , with r 1 = k x 1 k 2 . From x 1 and y 1 we create the Householder matrix H 1 = I - 2 u 1 u T 1 , u 1 = x 1 - y 1 k x 1 - y 1 k 2 . Then by direct calculation, you can check that A 2 = H 1 AH 1 = a 11 r 1 0 · · · 0 r 1 ˜ a 22 ˜ a 23 · · · ˜ a 2 N 0 ˜ a 32 ˜ a 33 · · · ˜ a 3 N . . . . . . . . . . . . . . . 0 ˜ a N 2 ˜ a N 3 · · · ˜ a NN . So applying H 1 on either side eliminates all but two entries in both the first row and first column of A . 22 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019 Subscribe to view the full document.

Notice also that the N × N matrix H 1 is orthnormal, as H T 1 H 1 = ( I - 2 u 1 u T 1 )( I - 2 u 1 u T 1 ) = I - 4 u 1 u T 1 + 4 u 1 u T 1 u 1 u T 1 = I , where the last equality follows since u T 1 u 1 = k u 1 k 2 2 = 1.  • Fall '08
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