Spectrum of an upper triangular matrix.

Question

Consider the algebra $A$ of upper triangular matrices. Given

$$a= \begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\0 & a_{22} & \dots & a_{2n}\\ 0 & 0 & \dots & a_{3n} \\ \vdots & \vdots & \dots & \vdots \\ 0 & 0 & \dots & a_{nn}\end{pmatrix}$$

I claim that the spectrum $\sigma(a) = \{a_{11}, a_{22}, \dots, a_{nn}\}$. I managed to show that $a_{11}, a_{22}, \dots, a_{nn}$ are all in the spectrum. Is there a way to show that the spectrum can not contain more elements?

Basically if $\lambda \notin \{a_{11}, \dots, a_{nn}\}$, my idea is that we can 'solve' the equation $(\lambda-a)b = b(\lambda-a) = I_n$ for some $b \in A$ by writing out the associated system of equations, but I want a cleaner less computational approach.

Any insight is appreciated!

Answer 1

One way to see this is the following (if you have heard about characteristic polynomials).

The characteristic polynomial of $a$ is the polynomial

$$ P(\lambda) = \det (\lambda I_n - a) = \prod_{k=1}^n(\lambda - a_{kk}) $$

where the second equality comes from the fact that the determinant of an upper triangular matrix is the product of its diagonal elements.

Now it is clear that the roots of $P$ are exactly $a_{11}, \dots, a_{nn}$ so $\sigma(a) = \{a_{11},\dots,a_{nn} \}$.