By Philip J. Davis
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1 Partition A as l — . A lr 11 12 kl (2. 1 . 2) k2 A = rr rl r2 where size A-^j = n-^ x nj The diagonal blocks A ^ square matrices of order n^. Example. X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X n 6 n. 2 are 1 is a symmetric partition of a 6 x 6 3 matrix. Square matrices are often built up, or compounded, of square blocks all of the same size. 18 Example Introductory Matrix Material x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x If a square matrix A of order nk is composed of n x n square submatrices all of order k, it is termed an (n, k) matrix.
1') /! V1! a: I The inverse permutation is designated by a 1 . Thus o- 1 (ik ) = k. , 0 's 0). 4) P = (a.. a. = /3 0 / otherwise. The ith row of P has a 1 in the a(i)th column and 0's elsewhere. The jth column of P has a 1 in the a ^(j)th row and 0's elsewhere. 6) P0A - (aa(i)ij). that is, P A is A with its rows permuted by a. 7) (x . 8) AP = (a , ). i,a 1 (j) That is, APa is A with its columns permuted by a 1 . 9) P p = p a t ax' where the product of the permutations a , from left to right. 11) (Pa)*Pa = P _lPa = Px = I.
Knopp, 1928, pp. s, (1-s,). 5) k s* = min(sk , 1 - sk ) = i ~ ' l - s, To this if w 0 < Sk - I' if ± 2 i Sk < ^ Different Values of s 11 nO° 2 Lemma. ¿k=1(sk - sk ) if and only if Ik= 1 sk < Proof. 2 0 < Sk _ Sk = Sk (1 “ Sk ) < and 1 r-iOO Hence Z,k= 2 sJ < other hand, if 0 < 00 implies sk i b rOO ¿ k=1 - min(Sk' 1-sk ) = sk - sk 2 (s^ - s^) < 00. (T,). 1). 6 for a more complete analysis. PROBLEMS 1. Let 2 = l/(k + 1) , k = 1, 2, ... Compute, An Introductory Geometrical Application 12 approximately, lim^^y (Tk )/y (T^) .
Circulant Matrices by Philip J. Davis