# Diagonalizable matrices

1. Dec 2, 2008

### lubricarret

1. The problem statement, all variables and given/known data

Let A and B be diagonalizable 2 x 2 matrices. If every eigenvector of A is an eigenvector of B show that AB = BA.

2. Relevant equations

D = PA(P^-1)

3. The attempt at a solution

Since the eigenvectors are equivalent, wouldn't it hold true that P_A = P_B?

If I have to show that AB = BA, I should be able to prove that

PAB(P^-1) = PBA(P^-1)

Since the eigenvectors of A are the eigenvectors of B, and
P = (Eigenvector_1, Eigenvector_2)

Then could I say that P_A = P_B, and (P^-1)_A = (P^-1)_B

and then cancel out P and (P^-1) from the equation PAB(P^-1) = PBA(P^-1) and then conclude that AB=BA?

Is my reasoning wrong here?

Thanks a lot!

2. Dec 3, 2008

### HallsofIvy

Staff Emeritus
Ouch! I get annoyed at having to open scanned papers- now I have to sit and watch you write it?

And you are using physics notation which gives me the pip.

If A and B are diagonalizable linear operators, then their eigenvectors form a basis for the vector space and written in that basis they are diagonal. If they have the same eigenvectors, then that single basis diagonalizes both! Yes, that is why you can use the same "P": there exist a single matrix P such that A= P-1DAP and such that B= P-1DBP where DA and DB are the appropriate diagonal matrices. What are AB and BA? Of course, diagonal matrices always commute.

By the way, this proof works for all dimensions, not just 2 by 2.

3. Dec 3, 2008

### lubricarret

Thanks HallsofIvy! I appreciate the explanation and for assuring me of that proof!!