Prior to reading the following post please consider the following notation and results.
- $\mathbf{F}$ is either $\mathbf{R}$ or $\mathbf{C}$
- $T\in\mathcal{L}(V)$ means the $T$ belongs to the set of all linear transformations over the vector space $V$.
- $\mathcal{M}(T)$ denotes the matrix of the linear transformation $T$.
- $E(\lambda,T) = \operatorname{null}(T-\lambda I)$ is the eigenspace of $\lambda\in\mathbf{F}$
$5.30$ If $T\in\mathcal{L}(V)$ has an upper triangular matrix with respect to some basis of $V$ then the invertibilty of $T$ is equivalent to all diagonal entries on the matrix of $T$ being non-zero.
I would like to know if the following Proof is correct?
Theorem. Given that $T\in\mathcal{L}(\mathbf{F}^5)$ and $\dim E(8,T) = 4$, then either $T-2I$ or $T-6I$ is invertible.
Proof. Assume on the contrary that both $T-2I$ and $T-6I$ are not invertible and let $v_1,v_2,v_3,v_4,v_5$ be the basis of $\mathbf{F}^5$ obtained by first extending the basis $v_1,v_2,v_3,v_4$ of $E(8,T)$.
Regardless of the image of $v_5$ under $T$ it is not difficult to see that $\mathcal{M}(T)$ is upper-triangular and by extension so is $\mathcal{M}(T-2I)$ and $\mathcal{M}(T-6I)$ then from theorem $\textbf{5.30}$ at least one of the entries on the diagonal of both $\mathcal{M}(T-2I)$ and $\mathcal{M}(T-6I)$ is zero but $(\mathcal{M}(T-2I))_{ii} = 6$ and $(\mathcal{M}(T-6I))_{ii} = 2$ for all $i\in\{1,2,3,4\}$ which implies that $(\mathcal{M}(T-2I))_{55} = (\mathcal{M}(T-6I))_{55} = 0$ but then $(\mathcal{M}(T))_{55} = 2$ and $(\mathcal{M}(T))_{55} = 6$ resulting in a contradiction.
$\blacksquare$
Your proof is correct. However, you made a mistake in the statement of theorem (5.30): it's not “all entries”; it's “all diagonal entries”. Besides, when you applied it, you only mentioned the diagonal entries.