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(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

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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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  • $\begingroup$ 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 $\endgroup$ – L G Jan 10 at 16:59
  • $\begingroup$ Yes: you also need $N + D = A$, and your proof should hold for any $A$. $\endgroup$ – user3482749 Jan 10 at 17:01
  • $\begingroup$ Right I'm not entirely sure where to start but I'll give it a go, thanks for pointing out my misunderstanding $\endgroup$ – L G Jan 10 at 17:03
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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!

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

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