In the Linear Algebra book by Hoffman Kunze, Exercise 12 of section 6.8 goes
If you thought about Exercise 11, think about it again, after you observe what Theorem 7 tells you about the diagonalizable and nilpotent parts of $T$.
The Exercise 11 that's mentioned in this question is
What is wrong with the following proof of Theorem 13? Suppose that the minimal polynomial for $T$ is a product of linear factors. Then, by Theorem 5, $T$ is triangulable. Let $\mathfrak B$ be an ordered basis such that $A=[T]_{\mathfrak B}$ is upper-triangular. Let $D$ be the diagonal matrix with diagonal entries $a_{11},\dots,a_{nn}$. Then $A=D+N$, where $N$ is strictly upper-triangular. Evidently $N$ is nilpotent.
The Theorem 13 here, as you might have already guessed, is
Let $T$ be a linear operator on the finite-dimensional vector space $V$ over the field $F$. Suppose the minimal polynomial of $T$ decomposes over $F$ into a product of linear polynomials. Then, there is a diagonalizable operator $D$ and a nilpotent operator $N$ on $V$ such that
1. $T=D+N$
2. $DN=ND$
This $D$ and $N$ are uniquely determined by 1 and 2 and each of them is a polynomial in $T$.
And Theorem 5 is
Let $V$ be a finite-dimensional vector space over the field $\mathfrak F$ and let $T$ be a linear operator on $V$. Then $T$ is triangulable if and only if the minimal polynomial for $T$ is a product of linear polynomials over $\mathfrak F$.
Now, the Theorem 7 mentioned in Exercise 12, as far as I understand, is
Let $V$ be a finite-dimensional vector space over the field $\mathscr F$ be a commuting family of triangulable linear operators on $V$. Then, there exists an ordered basis for $V$ such that every operator in $\mathscr F$ is represented by a triangular matrix in that basis.
Now that I have mentioned all the theorems, let me come back to the real question.
First of all, I don't understand properly what Hoffman Kunze means in that question (Exercise 12). Is it some silly typo that's there only in my edition, or is it there in all editions as a serious question? If it's the first one, please point the typo out. And if it's the later, please tell me what the question means. I guess, it somehow tries to connect Exercise 11 with Theorem 7. If that is right, then please give me some insight into how these two are related.
I tried searching online about this question, but couldn't find an website which didn't skip it. It's strange that no discussions about this problem have ever been done on MSE or Quora as well.
By the way, if you're interested, the solution of Exercise 11 is
We cannot guarantee that $D$ and $N$ commute, i.e., $DN=ND$ may not hold.
Edit: As Ben pointed out in the comments, maybe the authors want to suggest that if you can find the diagonalizable and nilpotent parts $D$ and $N$, then Theorem 7 says that you can simultaneously triangularize them. That's something I also thought about, but somehow this interpretation looks too simple to be on an exercise :|