Hoffman and Kunze Chapter 6 Theorem 13 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...
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.
$N^2=0$  where $N$ is a square matrix order $2$, $N$ will be similar  with which two matrices and how?