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

1. Jun 8, 2008

### mind0nmath

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?

2. Jun 8, 2008

### matt grime

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?

3. Jun 8, 2008

### mind0nmath

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

4. Jun 8, 2008

### mathwonk

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.

5. Jun 8, 2008

### mind0nmath

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?

6. Jun 9, 2008

### matt grime

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

7. Jun 10, 2008

### mind0nmath

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??

8. Jun 10, 2008

### morphism

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

9. Jun 10, 2008

### mind0nmath

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??

10. Jun 10, 2008

### morphism

If the scalar in the upper right corner is nonzero, then no, this matrix is not diagonalizable. It's been pointed out twice in this thread that a nonzero nilpotent matrix is not diagonalizable (general nxn matrices even, not just 2x2s) -- try to prove this.