# Abstract Linear Algebra: Eigenvalues & Eigenvectors

## Homework Statement

Let V be a finite dimensional vector space over ℂ . Show that any linear transformation T:V→V has at least one eigenvalue λ and an associated eigenvector v.

## The Attempt at a Solution

Hey everyone I've been doing sample questions in the build up to an exam and I came across this. Any help would be greatly appreciated as I'm struggling a bit.

Here is what I know:
• λ is an eigenvalue if there exists a non-zero vector v∈V such that Tv = λv.
• I also read this for complex: q(λ) = det (λI - T), where the zeros of q(λ) in ℂ are the eigenvalues of T.

What does the second point mean or how would I answer this properly. Thanks in advance.

## Answers and Replies

haruspex
Homework Helper
Gold Member
2020 Award
q(λ) = det (λI - T), where the zeros of q(λ) in ℂ are the eigenvalues of T.
That's also true for a vector space over R, every eigenvalue of A is a root of det (λI - A) = 0. (Can you see why?) But in R there might not be any real roots. Of course, you need to show the converse: that a root of the equation is necessarily an eigenvalue.

STEMucator
Homework Helper
In the complex plane, you are always guaranteed that there will be at least one eigenvalue for your transformation. This is assured by the fundamental theorem of algebra, which states that every polynomial has at least one root in ##\mathbb{C}##.

This is not true in ##\mathbb{R}## though, because the root of the characteristic polynomial might turn out to be complex.

Think about the matrix:

[0, -1
1, 0]

in the real numbers.