Invertible Linear transformation Eigenvalues I have read a thesis that claims:
A linear transformation is invertible iff 0 is not an eigenvalue of the representative matrix. 
I try to prove it, but couldn't, nor do I know if it's true. 
Can you please confirm or reject this thesis? 
Thanks, 
Alan
 A: Assuming you mean a linear operator, i.e. a linear transformation from a finite dimensional linear space $\;V\;$ to itself, then the claim is true:
$\;T:V\to V\;$ is invertible iff it is a bijection iff it is an injection iff $\;\ker T=\{0\}\;$ iff $\;Tv=0\cdot v=0\iff v=0\;$ iff zero is not an eigenvalue of $\;T\;$ .
A: first if $A$ is invertible, then $x \neq 0$ implies $Ax \neq 0.$ we can show this by contradiction. suppose $Ax = 0.$ multiplying on the left by $A^{-1}$ gives $x = 0$ contradicting the assumption $x \neq 0.$
we can show that $0$ is not an eigenvalue of $A.$ if $\lambda$ is an eignevlaue of $A,$ then there is an $x \neq 0$ such that $Ax = \lambda x.$ if $\lambda = 0$ then we would have $Ax = 0$ and $x \neq 0.$ that is not possible by previous remark.
now, we need to show that if $0$ is not an eigenvalue then $A$ is invertible. first we will show $x \to Ax$ is one-one. we will need that if $0$ is not an eignevalue, then $Ax = 0$ implies $x = 0$.  suppose $Ax = Ay.$ by linearity, $A(x-y) = 0.$ therefore $x-y = 0$ which in turn means $x = y.$  only thing left is that $x \to Ax$ is onto. i will have think of an elementary proof of that.
