p pa pb pc pd pe pf pg ph pi pk pl pm pn po pp pr ps pt pu pv pw px py

Перевод: permute

[глагол]
переставлять; менять порядок

Тезаурус:

1. (i) If A, B permute, so do powers of A, B. For example (ii) If A, B permute, and the eigenvalues of A are all different, then B can be expressed as a polynomial in A.
2. Now while, we can find a submatrix Y1 such that where E1 is of standard canonical form, Y1 is not of the general form D1 and so will not permute with C1.
3. But in this case, Y will, like D, permute with C; thus from (25) the modal matrix of A may be written XY, while (26) and (29) require so that B shares the modal matrix XY with A. Note that here, E need not be diagonal, so that B can be defective; but non-zero elements in the superdiagonal of E can only occur in a submatrix corresponding to a scalar submatrix in C.
4. Suppose, for example, that C has the leading submatrix then the most general submatrix of D which will permute with C1 is where d, e, f are arbitrary.
5. For example, if which we write as D = Diag(1,2,3), and then if C and D are to permute we must have so that d = e = f = g = h = k = 0, leaving C = Diag (a, b, c).
6. It follows that if C and D permute, so that DC = CD, then if .
7. Let d be an arbitrary diagonal matrix; then it will permute with the diagonal matrix Postmultiply (7) by d; then showing that Xd is a solution of (7).
8. Evidently A is the canonical form of A, and that of the defective matrix B. In this latter case, a postmultiplying non-singular matrix analogous to(9) which will permute with is where and are otherwise arbitrary.
9. Theorem XIV - If two matrices A, B permute, then provided at least one is non-defective, they share the same modal matrix.
10. In the light of the above, it is clear that two matrices having the same modal matrix do not necessarily permute; they will do so only if their spectral matrices permute.
11. Suppose we wish to permute the order of the bytes in a string.