# Diagonal entries of upper-triangular matrix contain ALL eigenvalues of an operator

This is 5.18 from Axler's Linear Algebra Done Right:

Theorem: Suppose T∈L(V) has an upper-triangular matrix with respect to some basis of V . Then the eigenvalues of T consist precisely of the entries on the diagonal of that upper-triangular matrix.

Proof:
Suppose ( v 1 ,…, v n ) (v1,…,vn) is a basis of V with respect to which T has an upper-triangular matrix where the diagonal entries are λ 1 ,…, λ n λ1,…,λn . Let λ∈F Then for matrix M(T−λI) where the diagonal entries are λ 1 −λ,… λ n −λ. λ1−λ,…λn−λ. We can suppose we are dealing with complex vector spaces. From 5.16 where have proven that T is not invertible iff one of the λ k λk 's equals 0 . Hence T−λI is not invertible if and only if λ equals one of the λj 's. In other words, λ is an eigenvalue of T if and only if λ equals one of the λj s, as desired.

Question: How does this prove that each of the eigenvalues of T is found in the diagonal? What if there are several distinct eigenvalues, but just one is found in the diagonal?

## 1 Answer

That cannot happen. What Axler does is this:\begin{align}\lambda\text{ is an eigenvalue}&\iff\det(M-\lambda\operatorname{Id})=0\\&\iff(\lambda_1-\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda)=0\\&\iff\lambda=\lambda_1\vee\lambda=\lambda_2\vee\cdots\vee\lambda=\lambda_n.\end{align}The second equivalence is where the fact that the matrix is triangular is used. Is there some step here that you do not understand?