If M is diagonalizable, find a matrix N^3 = M

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I have a question about diagonalizable matrix

So now I have showed M is a diagonalizable and I am asked to find a matrix N^3=M

Obviously if M=PDP^(-1)
then N=PD^(1/3)P^(-1) but I am wondering how to show it formally.
 
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Given that ##N=PD^{1/3}P^{-1}##, what is ##N^3=NNN##?
 
D H said:
Given that ##N=PD^{1/3}P^{-1}##, what is ##N^3=NNN##?
Strictly speaking, also need to show that this generates all solutions of the original equation.
 
haruspex said:
Strictly speaking, also need to show that this generates all solutions of the original equation.
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.
 
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.
True.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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