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.
Diagonalizable means that the matrix can be rewritten in diagonal form, where all the non-diagonal entries are equal to zero. This makes it easier to perform calculations and solve certain problems.
A matrix is diagonalizable if it has a full set of linearly independent eigenvectors. This means that the matrix can be transformed into diagonal form by using these eigenvectors.
If M is diagonalizable, it means that it can be written as M = PDP^-1, where P is a matrix of eigenvectors and D is a diagonal matrix with eigenvalues on the diagonal. To find N^3, we simply cube both sides of the equation, giving us N^3 = PD^3P^-1.
No, a non-square matrix cannot be diagonalizable because diagonalization requires finding a full set of linearly independent eigenvectors, which can only be done for square matrices.
Finding N^3 can help us perform calculations and solve problems involving M more easily. It can also help us understand the behavior and properties of M, as the diagonal form of a matrix can reveal important information about its eigenvalues and eigenvectors.