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By Yu B.J., Xu M.

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4) with these eigenvectors is non-singular so that A can be diagonalized by T. On the other hand, if some of the eigenvalues of A are degenerate, the eigenvalue problem need not provide n linearly independent eigenvectors to form a non-singular transformation matrix T . Before establishing the necessary and suf®cient condition for a matrix to be diagonalized by a similarity transformation we shall show the following fundamental theorem for a square matrix. 1. Every matrix satis®es its own characteristic equation.

Then the mutual orthogonalization of these two vector sets (like the Schmidt orthogonalization method) may lead to the desired diagonalization of A, as will be discussed in the next section. As a preparation we shall ®rst discuss the condition for the common submatrices T (í) to be non-singular. 4), the determinant of T” T is given by det T” T ˆ r ‰ (÷í ) ní det T (í) (2X5X6) íˆ1 Thus, if both T and T” are non-singular, then all T (í) , í ˆ 1, 2, F F F , r, are non-singular and vice versa. If one of T (í) is singular, at least one of T and T” must be singular.

7a) are arbitrary as long as the indices i belong to gÀ1 (í), any principal submatrices of ÷í T (í) are also positive de®nite; hence, their eigenvalues are all positive. Thus, their determinants (called the principal minors of ÷í T (í) ) are also positive: the ®rst few of them are ˆ ÷í T (í) jT ik j2 . 0; i P g À1 (í) ii ˆ 4 2 (÷í ) det k T (í) ii T (í) ij T (í) ji T (í) jj 5 (2X5X7b) . 0; À1 i, j P g (í) and so on. 7b) means that the signs of all diagonal elements of T (í) are given by the sign of ÷í .

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A Biordered Set Representation of Regular Semigroups by Yu B.J., Xu M.


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