Product of Diagonalizable matrices

  • Thread starter Thread starter Klungo
  • Start date Start date
  • Tags Tags
    Matrices Product
Klungo
Messages
135
Reaction score
1
Ignore post, I found a counterexample to (2).

I'm studying for an upcoming exam, and I'm a bit confused about how to go about proving or disproving the statement (2).

1.) Products of diagonalizable matrices are never diagonalizable.

I figured false and my counterexample is really just the square of the identity matrix. Since I is diagonalizable and I^2=I.

2.) Productions of diagonalizable matrices are always diagonalizable.

I'm not too sure if it's true to begin with. Trying to construct a counterexample for 2x2 matrices hasn't been successful yet. Any hints much appreciated.
 
Last edited:
Physics news on Phys.org
A matrix, A, on Rn, is "diagonalizable" if and only if there exist a basis for Rn consisting of eigenvectors of A. Can you use that?
 
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...
Back
Top