Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
Are you sure it doesn't say $D'$ and $N'$ commute with any polynomial in $T$? – Robert Israel Apr 10 '14 at 7:17
Oh yea it does. Typo! – mymindcastadrift Apr 10 '14 at 7:18
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.

share|cite|improve this answer
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 '14 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 '14 at 8:55

$N^2=0$ where $N$ is a square matrix order $2$, $N$ will be similar with which two matrices and how?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.