Question 15391
 Suppose V is an n dimensional vector space and that T is an operator on V. T has n distinct eigenvalues. Also suppose that S is also an operator on V which has the same eigenvectors as T ( not necessarily with the same eigenvalues).
Prove that ST = TS

Proof: If T has n distinct eigenvalues  {λi| i=1,2,..,n}(i.e. {{{lambda[i]}}} ) with corresponding eigenvectors {vi|i=1,2,..n}, 
then {vi|i=1,2,..n} is linear independent and so forms a basis of V. 
For each i, vi is an eigenvector of S means there exists scalar βi 
such that Svi = βi vi.
Consider ST (vi) = S(λi vi ) = λi S(vi) = λiβi vi and 
TS (vi ) = TS (βi vi ) = βi T(vi) =λiβi vi for every i=1,2,..n.
This shows ST = TS (since {vi|i=1,2,..n} is a basis of V).  

Fact. The eigenvectors {vi|i=1,2,..,n} of T is linearly independent 
 Proof: If {{{SIGMA}}} {{{c[i] v[i] = 0 }}} for some scalars ci, 
set L to be the product operator
 II(T –λj I) (j=1,2,..,n-1 ). Then, we have L(Σci vi) =Σci L(vi) 
= Σci (λi vi -λi vi) (i=1,2,..,n-1) + cn II(λn -λi )vn
= cn II(λn -λi )vn  = 0 .
Since all {&#955;i} are distinct, II(&#955;n -&#955;i ) <> 0. (i=1,2,..,n-1) 
We obtain cn = 0. Similarly, we can prove that all other scalars ci are zeros. This shows that {vi} are independent. 

 Kenny
 PS. I don't quite believe that some questions like this would be posted
 here.