Question 5452: Prove that a linear operator T on a finite-dimensional vector space is invertible if and only if zero is not an eigenvalue of T.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Prove that a linear operator T on a finite-dimensional vector space is invertible if and only if zero is not an eigenvalue of T.
Proof: ==> If T in L(V,V) is invertible, and if c is an eignevalue of
of T,then there exists nonzero vector x in V such that
Tx = c x. Suppose c = 0,then Tx = 0 implies Ker(T) is not
{0} and so T is not innvertible. This leads a contradiction.
<== if zero is not an eigenvalue of T, then for any nonzero vector
x in V, Tx cannot be 0. This means Ker(T) = {0} and so T is
one-to-one. Hence,T is invertible.
Anotherway, use the fact that an eigenvalue is a root of the charcteristic ploynomial det(cI -T) = 0 . c = 0 <--> det(T) =0 <--> T is singular
<--> T is not invertible.
Kenny
|
|
|
