Diagonalizable Matrices and Commutativity: Proving AB = BA

[lubricarret]

Homework Statement

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

Homework Equations

D = PA(P^-1)

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!

 

[HallsofIvy]

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^-1D_AP and such that B= P^-1D_BP where D_A and D_B 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.

 

[lubricarret]

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