Source: Linear Algebra by David Lay (4 edn 2011). p. 269 Theorem 5.1.1.
For simplicity, consider the $3\times 3$ case. If $A$ is upper triangular, then
$ A-\lambda I=
\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
0 & a_{22} & a_{23}\\
0 & 0 & a_{33}
\end{bmatrix}
–
\begin{bmatrix}
\lambda & 0 & 0\\
0 & \lambda & 0\\
0 & 0 & \lambda
\end{bmatrix}
=\begin{bmatrix}
a_{11}-\lambda & a_{12} & a_{13}\\
0 & a_{22}-\lambda & a_{23}\\
0 & 0 & a_{33}-\lambda \\
\end{bmatrix}$$\lambda$ is an eigenvalue of $A \iff$ The equation $(A-\lambda I)x=0$ has a
nontrivial solution. $\iff$ $(A-\lambda I)x=0 $ has a free variable $ \iff $
Because of the zero entries in $A-\lambda I$, at least one of the entries on the diagonal of $A-\lambda I$ is zero.
$\color{red}{\iff} \lambda$ equals $\color{red}{one \, of } $ the entries $a_{11},\ a_{22},\ a_{33}$ in $A$.
For the lower triangular case, see Question 5.1.28.
$1.$ I don't understand the red $\color{red}{\iff}$. $a_{11} – \lambda = 0 \iff$ The first column is $\mathbf{0}. \iff$ $x_1$ is a free variable. But what about the other entries?
$2.$ I linked an analogous question. How does the proof overhead proves that all of the eigenvalues = all its diagonal entries, when it states $\color{red}{one \, of } $?
Best Answer
It's equivalent to showing that an upper triangular matrix is injective iff none of its diagonal entries are $0$.
If one of the diagonal entries, say the $i$th, is $0$, then restricting the operator to the span of the first $i$ basis vectors gives a map into the span of the first $i-1$ basis vectors, which by Rank-Nullity cannot be injective.
Now, if all of the diagonal entries are nonzero, then the columns must be independent: the last column is the only one with a nonzero $n$th entry, so its coefficient has to be $0$. Continuing this way, you can kill all the coefficients.