For a square matrix A over $\mathbb{C}$, Proofs that matrices D and N exist with A=D+N under different conditions

Question

(i) D is Diagonalizable

This one i believe to be fairly straightforward, if D is diagonalizable then we can allow $D^t = I$ (where I is the identity) and therefore D id diagonalizable and therefore A=N+D is straightforward.

(ii)N is Nilpotent

So when there exists an r such that $N^r=0$

(iii)DN = ND

Each one seems rather trivial, e.g for (ii) D could be any square matrix of the same size as N and it be a Square matrix A and for (iii) you just prove commutativity for matrices?

I believe I'm missing something quite important here but can't figure out what

Answer 1

First: I'm very sure that you've entirely misunderstood the question, and you need to find $D$ and $N$ such that all of these conditions hold at once. I'll edit a proof of that question into this answer later today, if nobody beats me to it.

This one i believe to be fairly straightforward, if D is diagonalizable then we can allow Dt=I (where I is the identity) and therefore D id diagonalizable and therefore A=N+D is straightforward.

I have no idea what you mean here, but if you're only trying to find $D$ satisfying this condition, just take $D$ to be the identity.

(ii)N is Nilpotent So when there exists an r such that $N^r=0$

Again, if you only wanted this, just take $N = 0$.

Each one seems rather trivial, e.g for (ii) D could be any square matrix of the same size as N and it be a Square matrix A and for (iii) you just prove commutativity for matrices?

You could try, but matrix multiplication isn't commutative, so you would fail.

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Answer 2

What you are probably missing here is that all conditions should hold simultanously. You can not choose D=I because than A-D=N would not necessarily be nilpotent. Also, commutativity for matrices is not something that always holds!