matrix question

[fallow heart]

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[fallow mirage]

That isn’t true you can take for example two nilpotent matrices of nilpotent indicator 2 A and B you have the equality (A+B)^3=A^3+3A^2B+3AB^2+B^3=0 but they don’t necessarily commute

like for example $A=\begin{pmatrix} 0& 1 \ 0 & 0 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix}$

[lavish moon]

Rotor

[fallow mirage]

both don’t commute and are nilpotent

And you have (A+B)^3 which is non zero

But I have an interesting problem if A and B are two matrices such that A+B=AB prove that A and B commute

I’ll give a hint consider (I-A)(I-B) here « I » is the identity matrix

[lapis sparrow]

@pale kraken has given 1 rep to @fallow mirage

[fallow mirage]

Yep

Good job

You are very welcome

Sure

Also this only works because a matrice and it’s inverse commute

Wait I think my example wasn’t good

(A+B)^3 is not zero

My bad

But I’m sure it’s possible to find two nilpotent matrices that don’t commute such that their sum is nilpotent

In dimension 3 for example

[fallow mirage]

a way you can do it is just do the calculations of course

Lemme see if there is another way though

[sly wyvern]

order 3 determinant

It will be commutative

It is the other way around

You are right

AB should be equal to BA