y N 1 y L 1 y L y L N 2 If we recover x from y using least squares b x A y A T

# Y n 1 y l 1 y l y l n 2 if we recover x from y using

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. . y [ N - 1] . . . y [ L - 1] y [ L ] . . . y [ L + N - 2] If we recover x from y using least-squares, b x = A y = ( A T A ) - 1 A T y , then the N × N system we need to invert, H = A T A is also Toeplitz (and is of course symmetric and non-negative definite): H = h 0 h 1 · · · h N - 1 h 1 h 0 h 1 · · · h N - 2 h 2 h 1 h 0 · · · h N - 3 . . . h N - 1 · · · · · · h 0 . 2 We are going to save some space in this section by using subscript notation to index signals that are going in a matrix; i.e. a k instead of a [ k ]. 17 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019

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Here is a quick example in MATLAB: >> A = toeplitz([1; randn(5,1); zeros(3,1)], [1 zeros(1,3)]) A = 1.0000 0 0 0 -0.4336 1.0000 0 0 0.3426 -0.4336 1.0000 0 3.5784 0.3426 -0.4336 1.0000 2.7694 3.5784 0.3426 -0.4336 -1.3499 2.7694 3.5784 0.3426 0 -1.3499 2.7694 3.5784 0 0 -1.3499 2.7694 0 0 0 -1.3499 >> H = A’*A H = 23.6023 6.8156 -5.0905 1.9151 6.8156 23.6023 6.8156 -5.0905 -5.0905 6.8156 23.6023 6.8156 1.9151 -5.0905 6.8156 23.6023 Symmetric Toeplitz systems appear frequently in linear prediction, array processing, adaptive filtering, and other areas of statistical sig- nal processing. An N × N Toeplitz system H can be inverted in O ( N 2 ) time using the Levinson-Durbin algorithm. The algo- rithm is relatively easy to derive, and even easier to implement. The increase in efficiency it offers is significant, as the difference between O ( N 3 ), the cost of solving the system using a general linear solver, and O ( N 2 ) is enormous even for moderate N . 18 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019
Technical Details: LU , Cholesky, and Symmetric QR Factorizations Example of LU factorization Start with the original matrix A = [ r ]2 1 - 1 - 3 - 1 2 - 2 1 2 Eliminate the lower-left term. In this case we can add the first row to the third row, i.e. [ r ] 1 0 0 0 1 0 1 0 1 [ r ]2 1 - 1 - 3 - 1 2 - 2 1 2 = [ r ] 2 1 - 1 - 3 - 1 2 0 2 1 L 1 A = A 1 Eliminate the term in (row,column) = (2 , 1) by adding 3 / 2 the top row to the second row: [ r ]1 0 0 1 . 5 1 0 0 0 1 [ r ] 2 1 - 1 - 3 - 1 2 0 2 1 = [ r ] 2 1 - 1 0 0 . 5 0 . 5 0 2 1 L 2 A 1 = A 2 Finally, we eliminate (3 , 2) by subtracting 4 times the second row from the third row: [ r ]1 0 0 0 1 0 0 - 4 1 [ r ] 2 1 - 1 0 0 . 5 0 . 5 0 2 1 = [ r ] 2 1 - 1 0 0 . 5 0 . 5 0 0 - 1 L 3 A 2 = U . 19 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019

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So we have: L 3 L 2 L 1 A = U , where U is upper triangular and the L i are all lower triangular with 1 along the diagonal and exactly 1 non-zero off-diagonal term. Using the facts that 1. the inverses of L i are also lower triangular with 1 down the diagonal and exactly one non-zero off-diagonal term (show this at home!), and 2. the product of two lower triangular matrices with 1 down the diagonal is again lower-triangular with 1 down the diagonal (show this at home!), we have A = ( L - 1 1 L - 1 2 L - 1 3 ) U = LU , where L is lower diagonal with 1 down the diagonal. In the example above, we have L = L - 1 1 L - 1 2 L - 1 3 = [ r ] 1 0 0 0 1 0 - 1 0 1 [ r ]1 0 0 - 1 . 5 1 0 0 0 1 [ r ]1 0 0 0 1 0 0 4 1 = [ r ]1 0 0 - 1 . 5 1 0 - 1 4 1
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