Question

In the book chapter6.2, it says when A and B both diagonalizable, AB=BA is true if and only if they have the same eigenvectors. It's easy to obtain that AB=BA,but how to prove vise versa??

Answer 1

The statement does not depend of basis choice, so let's consider the basis, consisting from eigenvectors of A. Matrix presentation of A is diagonal in that basis. AB=BA implies that matrix presentation of B is a diagonal matrix, which means that basis vectors are eigenvectors of B.

Answer 2

Let's say if A has totally different eigenvalues, ΛB=BΛ (Λ , the diagonal matrix of A) deduces that B is also diagonal, very well. What if Λ has the same eigenvalue, I don't think B is diagonal deduced from ΛB=BΛ.
Also, AB=BA equal to ΛB=BΛ? IF A=SΛS^(-1), then SΛS^(-1)B=BSΛS^(-1), how I get ΛB=BΛ?

Answer 3

I like your question. Generally speaking the statement of your theorem is not true. However, existence of common basis formed by eigenvectors of diagonalizable A and B is equivalent to AB = BA. Consider extreme case when all eigenvalues of A are the same. Then any vector of n-dimensional space Rn is an eigenvector of A. In that case the statement that A and B have common basis formed by eigenvectors is trivial.
I will show proof of the general case later today.

Answer 4

Your last question:
"Also, AB=BA equal to ΛB=BΛ? IF A=SΛS^(-1), then SΛS^(-1)B=BSΛS^(-1), how I get ΛB=BΛ?"

Answer:
When A = S*DIAG*S^(-1) then B transforms to B' = S*B*S^(-1), so AB=BA => DIAG*B' = B'*DIAG. In other words, property AB = BA does not depend from choice of the basis.

Answer 5

Thanks for your patient answer! There's still doubts..but I'll wait and see your proof of general case first.

Answer 6

General case.
1. There exists basis formed by eigenvectors of A, since A is diagonalizable.

2. Sort the basis vectors by ascending order of corresponding eigenvalues.
In that basis:
a) Matrix representation of A is a diagonal matrix DIAG.
b) Matrix representation of B is cellular-diagonal matrix CELL, as follows from equation AB=BA.
3. Since B-cells outside of main diagonal are zeroes, each group of basis vectors, corresponding the same eigenvalue of A, spans a B-invariant subspace.
4. Restriction of B on that subspace is also diagonalizable.
5. There exists a basis of that subspace formed by eigenvectors of B.
6. All base vectors of this basis are eigenvectors of A, since any vector of this subspace is an eigenvector of A ( as in extreme case above).
7. Union of all such base systems forms common basis in Rn, consisting of common eigenvectors of A and B.