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

(a) Prove that a linear operator $T$ on a finite-dimensional vector space is invertible if and only if zero is not an eigenvalue of $T$.

(b) Let $T$ be an invertible linear operator. Prove that a scalar $\lambda$ is an eigenvalue of $T$ if and only if $\lambda^{-1}$ is an eigenvalue of $T^{-1}$.

MY ATTEMPT

(a) Let us suppose that $\lambda = 0$ is an eigenvalue of $T$. Thus $\det([T]_{\mathcal{B}}) =\det([T]_{\mathcal{B}} - 0 I_{n}) = 0$, and $T$ is not invertible.

Conversely, let us suppose that $T$ is not invertible. Then $0 = \det([T]_{\mathcal{B}}) = \det([T]_{\mathcal{B}} - 0I_{n})$, and $0$ is an eigenvalue of $T$.

(b) Indeed, provided that $\lambda \neq 0$, one has \begin{align*} Tv = \lambda v \Rightarrow T^{-1}(Tv) = T^{-1}(\lambda v) \Rightarrow (T^{-1}T)v = \lambda(T^{-1}v) \Rightarrow T^{-1}v = \lambda^{-1}v \end{align*}

Once again, I am mainly interested in the wording of my proof. Any comments or contributions are appreciated.

Answer 1

Your proofs are correct.

I would edit the second assertion to put the equations on separate lines separated by the implications, rather than running them all together.

I think that a proof that relies more directly on the concepts and avoids determinants would be better. If $0$ is an eigenvalue then there is a nonzero $v$ with $Tv=0$ so T is not injective, so it's not invertible. Conversely, if $T$ is not invertible then it's not bijective. For a linear transformation from a finite dimensional space to itself, bijective is equivalent to injective. But an noninjective linear transformation has a kernel, so has $0$ for an eigenvalue.