Proof that Eigenvalues are the Diagonal Entries of the Upper-Triangular Matrix in Axler

Solution 1:

I understand why this idiom (which is common in math writing) might seem confusing, but what the author is saying is correct. When he says that

$\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ equals one of the $\lambda_j$'s

he means that

$\lambda$ is an eigenvalue of $T$ if and only if $\lambda\in\{\lambda_1,\ldots,\lambda_n\}$

or, to phrase it another way,

$\lambda$ is an eigenvalue of $T$ if and only if $\lambda=\lambda_1$, or $\lambda=\lambda_2$, ..., or $\lambda=\lambda_n$

Thus, if I set $\lambda$ equal to $\lambda_1$, the right side of the biconditional is true, so that $\lambda$ is an eigenvalue of $T$ when $\lambda=\lambda_1$; and similarly with all of the diagonal entries $\lambda_1,\ldots,\lambda_n$.