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?
Best 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?