Question 5373
 Let T be a linear operator on a finite-dimensional vector space V, and let
b (beta) be an ordered basis for V. Prove that l (lamda) is an eigenvalue of T if and only if l (lamda) is an eigenvalue of [T]b (beta).

 ==>  if lamda is an eigenvalue of T , then there exists nonzero vector 
      x in V such that Tx = lambda x,
      Since B(better using capital letter for basis) is an order basis
      of V, let B ={v1,v2,..,vn} (assume dim  V = n)
      x = E aivi for scalars a1,a2,..,an (E means summation over i)
      i.e. xB = (a1,a2,..,an)^T (T means transpose) be a column vector(in F^n)
      Let [T]B = [Tij](nxn matrix)
      Since Tx = lambda x, expressing in the o.b. B, we
      have [Tx]B = [lambda x]B, 
   So, we get [Tx]B =[T]B xB = lambda xB
   Also, x is nonzero vector implies xB is non-zero in F^n.
   This shows lambda is an eigenvector of [T]B.
 <== If lambda is an eigenvector of [T]B , then
     there exists a non-zero column vector w=(c1,c2,..,cn)^T in F^n
      such that [Tx]B w = lambda w.
     Note B = {v1,v2,..,vn} and set x = E civi
     then we have xB = w and 
     [Tx]B = [T]B xB = [T]B w = lambda w = lambda xB,
     This implies Tx = lambda x because B is a basis of V.
   (a linear operator isnquely determined by the values on a basis)
     Also, x is non-zero since w is non-zero.
     This proves lambda is an eigenvalue of T.

 Kenny

 PS:
  By definition of [T]B = [Tij] (nxn matrix) 
  if B = {v1,v2,..,vn}
   for each i,T (vi) = E Tij vj (summation over j)