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I am completely stumped at the proof of Theorem 13 in Chapter 6, Hoffman and Kunze. The theorem goes:


Let $T$ be a linear operator on the finite-dimensional vector space $V$ over field $F$. Suppose that the minimal polynomial for $T$ decomposes over $F$ into a product of linear polynomials. Then there is a diagonalizable operator $D$ on $V$ and a nilpotent operator $N$ on $V$ such that

(i) $T = D + N$

(ii) $DN = ND$

The diagonalizable operator $D$ and the nilpotent operator are uniquely determined by (i) and (ii) and each of them is a polynomial in $T$.


The crux was to prove the uniqueness. In the proof given, there was a sentence: "$D'$ and $N'$ commute with any polynomial in $T$; hence they commute with $D$ and with $N$"

Does this mean $DD' = D'D$ and $NN' = N'N$? How did this come about? I am definitely overlooking something obvious...

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Are you sure it doesn't say $D'$ and $N'$ commute with any polynomial in $T$? –  Robert Israel Apr 10 at 7:17
    
Oh yea it does. Typo! –  mymindcastadrift Apr 10 at 7:18

1 Answer 1

up vote 0 down vote accepted

First a comment: Uniqueness is fairly routine; the crux is in the existence which requires Chinese Remainder Theorem.

Now to your question: yes, $DD=D'D$ and $NN'=N'N$. The theorem constructs $D$ and $N$ as polynomials $T$. So, if we have possible second candidate, then two polynomials in $T$ which give $D$ and $D'$ will commute.

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Oh okay I see. The book constructed the existence of D and N by looking at factors of minimal polynomials. This implied that both D and N are polynomials in T. So the reasoning was actually just one step from the properties of a polynomial. Thanks! Just curious: How does the Chinese Remainder Theorem get involved? –  mymindcastadrift Apr 10 at 7:24
    
I haven't read that book; but if you are looking for a polynomial satisfying some congruence conditions you are appealing to that theorem in the ring $\mathbf{C}[X]$. –  P Vanchinathan Apr 10 at 8:55

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