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

(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

• I'm not sure I understand. Are you saying we need to satisfy all three conditions simultaneously? – Joe Jan 10 at 16:48
• @Joe I misunderstood the question i believed it would be seperate but it means simultaneously, still not sure how to do the question though haha – L G Jan 10 at 17:06
• – Viktor Glombik Jan 10 at 18:26

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.

• I think you're right thank you, forgot about commutivity as well oops! Just at a quick glance would it be wrong to say N = 0 and D = I as then ND = 0 = DN, and a single example would suffice to answer the question – L G Jan 10 at 16:59
• Yes: you also need $N + D = A$, and your proof should hold for any $A$. – user3482749 Jan 10 at 17:01
• Right I'm not entirely sure where to start but I'll give it a go, thanks for pointing out my misunderstanding – L G Jan 10 at 17:03

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!

• Yes you're right, i believe i have and that's also a schoolboy error i've made! Thanks! – L G Jan 10 at 17:00