Abstract Linear Algebra: Eigenvalues & Eigenvectors

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SUMMARY

The discussion focuses on the existence of eigenvalues and eigenvectors for linear transformations in finite-dimensional vector spaces over the complex numbers. It establishes that any linear transformation T: V → V has at least one eigenvalue λ and an associated eigenvector v, supported by the characteristic polynomial q(λ) = det(λI - T). The fundamental theorem of algebra guarantees that this polynomial has at least one root in ℂ, confirming the presence of eigenvalues. In contrast, the discussion highlights that while every eigenvalue of a matrix A in ℝ corresponds to a root of det(λI - A) = 0, real roots may not exist.

PREREQUISITES
  • Understanding of linear transformations and vector spaces
  • Familiarity with eigenvalues and eigenvectors
  • Knowledge of characteristic polynomials and determinants
  • Basic concepts of complex numbers and the fundamental theorem of algebra
NEXT STEPS
  • Study the properties of eigenvalues and eigenvectors in linear algebra
  • Learn about the characteristic polynomial and its applications in determining eigenvalues
  • Explore the differences between eigenvalues in complex and real vector spaces
  • Investigate the implications of the fundamental theorem of algebra on polynomial equations
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Students studying linear algebra, mathematicians focusing on eigenvalue problems, and educators teaching concepts related to vector spaces and linear transformations.

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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.

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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.
 
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teme92 said:
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.
 
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.
 

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