Let [PHI] be a right quaternion linear map from [M.sub.2](H) into itself Then [[sigma].sub.l](A) = [[sigma].sub.l]([PHI](A)) for all A [member of] [M.sub.2](H) if and only if there exists an

invertible matrix B [member of] [M.sub.2](R) such that [PHI](A) = BA[B.sup.-1].

On the other hand, if [chi] is the inner automorphisms induced by the

invertible matrix Q = I + [e.sub.ri], as above [chi](a)x[chi](q) + [chi](c)x[chi](b) = 0, for all x [member of] R.

The restrictions in the form of (3) and (4) are triangular if and only if there exists an

invertible matrix [P.sub.1] such that the matrix [P.sub.1]f ([A.sub.0], [A.sub.+]) [P.sub.0] is lower triangular, where

DO generates an

invertible matrix M as the encryption key to encrypt [[??].sub.i], such that [p'.sub.i] = [[??].sub.i]M.

(i)[??](ii) Premultiply the matrix (A,C) with the

invertible matrix ([bar.A], [A.sub.[perpendicular to]]) to get the identity

Robinson, "Covariance of Moore-Penrose inverses with respect to an

invertible matrix," Linear Algebra and its Applications, vol.

such that [??](z) = U([z.sup.m])P(z)[U.sup.-1](z) with P(z) taking the form (33), where [U.sup.0] is a constant

invertible matrix. It is evident that

If D is an

invertible matrix, then it has no zero eigen-values (c = 0 in (9) and, thus, for all i, [absolute value of 1 - [lambda][d.sub.i]] < 1.

In order to derive an SP for Kaczmarz relaxation (2.8), we shall consider a general

invertible matrix A (not necessarily SPD), b [member of] [R.sup.n] a given vector and the linear system

(1a) X ([A.sub.0](k)C, [A.sub.+] (k)C) = X ([A.sub.0](k) ,[A.sub.+] (k))C, for every

invertible matrix C.

Note that this matrix is unimodular and hence unimodularly equivalent to a constant,

invertible matrix. The program yields no column-reduced R for b [less than or equal to] 4.

A G -trapdoor for A is a matrix R[member of][Z.sup.(m-w)xw] such that [mathematical expression not reproducible] for some

invertible matrix [mathematical expression not reproducible].