# Way to generalize the relationship between T and T^2?

Hey,
I know that if T(linear transformation in finite dim-vector space) is diagonalizable, then the matrix A that represent T is diagonalizable if there exist a matrix P=! 0 that is invertible and A=P^-1 * D * P for a diagonal matrix D.
I also know that raising A to a positive power will equal the right side with D raised to that same power. i.e. A^n = P^-1 * D^n * P
Does this imply that if T is diagonalizable so is T^2? what if T^2 is diagonalizable, then does that imply T is diagonalizable?

I think that it is an iff statement. Can anyone shine some light into this situation? maybe a counter example? is there a way to generalize the relationship between T and T^2?

matt grime
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T diagonalizable certainly means T^2 is diagonalizable. The reverse implication is false, in general. To see why: if t is an e-value of T, then t^2 is an e-value of T^2. Now, what if s is an e-value of T^2? We're surely going to want sqrt(s) an e-value of T, but what if the matrices are over R and s=-1?

So if T is over the complex numbers, then both directions are true?
i.e. T is diagonalizable <=> T^2 is diagonalizable?

mathwonk
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2020 Award
no. the standard example of a on diagonalizable matrix is a nilpotent matrix. so if T^2 = 0, then T^2 is obviously diagonal but T may not be.

e.g. differentiation of linear functions is square nilpotent, but there are no linear functions such that Df = af except f =0.

I thought for T to be diagonalizable, it should have dimT distinct eigenvalues or dim(T) linearly independent eigenvectors. Then how is the Zero matrix diagonalizable? aren't the roots all the same(i.e. =0)? and null(T)=C^n?

matt grime
Homework Helper
It is diagonalizable by your own criteria: any basis is a set of linearly independent eigenvalues.

How about the addition of two diagonalizable matricies? Is it always diagonalizable in some basis if the field is the complex numbers?
my initial thought is: yes. by the spectral theorem. but not 100% sure since I suspect we might not be able to find a basis for the addition of the matrices which in turn would make it diagonalizable.
any thoughts??

morphism
Homework Helper
Have you attempted to find a counterexample? Keep in mind that a nonzero nilpotent matrix is not diagonalizable.

Here's a new question(related to the subject). Is the 2x2 matrix with only a scalar in the right upper corner and all other entries = 0, diagonalizable? That matrix, call it T, is nilpotent. T^2 is the zero matrix. Isn't the zero matrix diagonalizable? it is in the diagonal form so it must be. Is T diagonalizable??

morphism