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).
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! 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)
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