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: $\Rightarrow$: let $v \in V$, then $T(v)=\lambda v$. Since T is invertible, it is injective, so by definition, T(v)=0,
Question: I am not sure what to argue after that, some definitions to think about?
$\Leftarrow$: If zero is not an eigenvalue of T,then...
Question: I was told to think of $T-\lambda I$, a linear mapping. But not sure how to utilize.
Thanks@
One doesn't need to use determinants or look too closely at $T - \lambda I$ to address this, to wit:
Let
$T \in \mathcal L(V), \tag 0$
with
$\dim V < \infty, \tag{0.5}$
and suppose that $0$ is not an eigenvalue of $T$; then $T$ is injective; for if
$v_1, v_2 \in V \tag 1$
with
$Tv_1 = Tv_2, \tag 2$
then
$T(v_1 - v_2) = Tv_1 - Tv_2 = 0 = 0(v_1 - v_2); \tag 3$
thus if
$v_1 - v_2 \ne 0, \tag 4$
$v_1 - v_2$ is an eigenvector of $T$ with eigenvalue $0$, contrary to hypothesis; therefore
$v_1 - v_2 = 0 \Longrightarrow v_1 = v_2, \tag 5$
establishing the injectivity of $T$; in light of (0.5) we see that $T$ is also surjective; hence $T$ is a linear isomorphism of $V$, and is invertible.
We next assume that $T$ is invertible, and that
$Tv = \lambda v, \; v \ne 0; \tag 6$
in this event we may write
$0 \ne v = Iv = T^{-1}Tv = T^{-1} (\lambda v) = \lambda T^{-1}v \tag 7$
then clearly
$\lambda \ne 0, \tag 8$
and we are done.