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

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

Theorem: Suppose $$T \in L(V)$$ 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 $$(v_1, \ldots , v_n)$$ is a basis of $$V$$ with respect to which $$T$$ has an upper-triangular matrix where the diagonal entries are $$\lambda_1, \ldots, \lambda_n$$.

Let $$\lambda \in F$$

Then for matrix $$M(T - \lambda I$$) where the diagonal entries are $$\lambda_1 - \lambda, \ldots \lambda_n - \lambda.$$ 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 $$\lambda_k$$'s equals $$0$$. Hence $$T - λI$$ is not invertible if and only if $$λ$$ equals one of the $$λ_j$$'s. In other words, $$λ$$ is an eigenvalue of $$T$$ if and only if $$λ$$ equals one of the $$λ_j$$s, as desired.

Question:

This only showed that one of the diagonal entries is en eigenvalue but not all of them as the theorem claimed.

• $\lambda$ is arbitrary, as long as it is one of the $\lambda_j$. It shows one of them, but it is any one of them, thus they are all the eigenvalues. Dec 25, 2012 at 15:38
• Determinant of lower/upper triangular matrix is just product of it’s diagonal entries. (It can be observed from definition/property of determinant that involves algebraic minors) Apr 30, 2022 at 18:17

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 $\lambda_1,\ldots,\lambda_n$.

• Would you please explain how $\lambda = \lambda_1$ and $\lambda = \lambda_2$ and $\lambda = \lambda_i$ for all $1 \le i \le n$? I still don't apprehend this idea?
– user53259
May 26, 2014 at 16:30