The discussion revolves around finding a matrix \( N \) such that \( N^3 = M \), where \( M \) is given to be diagonalizable. The original poster has established that \( M \) is diagonalizable and is exploring the implications of this for determining \( N \).
Participants are actively engaging with the problem, with some providing guidance on the necessity of showing that \( D^{1/3} \) exists and that \( (D^{1/3})^3 = D \). There is a recognition of the need to clarify the scope of the solution being sought.
There is an emphasis on the distinction between finding a specific matrix \( N \) and generating all matrices \( N \) such that \( N^3 = M \). The discussion also highlights the importance of confirming the existence of the cube root of the diagonal matrix \( D \).
Strictly speaking, also need to show that this generates all solutions of the original equation.D H said:Given that ##N=PD^{1/3}P^{-1}##, what is ##N^3=NNN##?
I'm not sure what you mean by that. Per the title of the thread, victoranderson has to find a matrix N such that N^3=M, not all matrices N such that N^3=M.haruspex said:Strictly speaking, also need to show that this generates all solutions of the original equation.
True.D H said:I'm not sure what you mean by that. Per the title of the thread, victoranderson has to find a matrix N such that N^3=M, not all matrices N such that N^3=M.
What he does need to show is that his D^(1/3) always exists and that (D^(1/3))^3=D.