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.

 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