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By A Seidenberg

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ZB/k 1 /: By passing to determinants, we deduce that P is a divisor of 1 in CŒX, so it is indeed constant. 11. y/ D det k X xi iD0 2 D det 4 k X xi iD0 iŠ ! 0 Ai @ Ai 1 0 k j X y jA A det @ jŠ jD0 1 3 2 0 k X xi yj X A Ap 5 @ Aj A5 D det 4 jŠ iŠjŠ pD0 iCjDp k X yj jD0 iŠ ! x/P. 0/ D 1; so P is a constant polynomial, because it does not have complex roots. 0/ D 1: We can also proceed in a different way. According to the hypotheses, A is nilpotent. The sum of nilpotent matrices which commute with each other is a nilpotent matrix.

Denote by ˚ Zp D 0; 1; : : : ; p A1o ; Z n « 1 ; Zq D e 0; e 1; : : : ; q pq 11o n D b 0; b 1; : : : ; pq the remainder (or residue) class sets relative to p; q;, respectively pq: Remember that, for any positive integer m, we can define (on the set Z of the integers) the © Springer Science+Business Media LLC 2017 T. 1007/978-0-8176-4629-5_2 25 26 2 Cardinality relation of congruence modulo m by a Á b mod m if and only if a b is divisible by m (or, equivalently, if a and b give equal remainders when divided by m).

F/, such that A D PBP 1 . In this case, we say that A and B are similar. XIn A/ the characteristic polynomial of f . It is easy to see that f does not depend on A, because two similar matrices have the same determinant. Before passing to some problems, let us point out some other useful facts. C/ and A is invertible, then AB and BA are similar. BA/A 1 . Thus AB D BA in this case. C/. A C XIn / is a nonzero polynomial. Actually, for the same reason, there is such a sequence "k with lim "k D 0. zIn BAk /: It is enough to make k !

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