SOLUTION: 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.

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
RELATED QUESTIONS
Let T be a linear operator on a finite-dimensional vector space V, and let b (beta) be... (answered by khwang)

Suppose V is an n dimensional vector space and that T is an operator on V. T has n... (answered by khwang)

Let "T" be a linear Operator on a finite dimensional space "V" and let "c" be a Scalar. (answered by khwang)

Let T be a linear operator in V. Prove that T^2 = 0 if and only if the range T is a... (answered by khwang)

For the following linear operator T on the vector space V, test T for diagonalizability,... (answered by khwang)

suppose that V is finite-dimensional space and T:V->V is a linear operator. 1. prove... (answered by ikleyn)

Discuss in detail why the dimension of a finite dimensional vector space is... (answered by ikleyn)

Let X be the design matrix of a linear regression problem with m rows (samples) and d... (answered by asinus)

Let T, U: V to W be linear transformations. Prove that: a. R(T+U) is a subset of... (answered by khwang)