n x n matrix?

Question

if A is an nxn matrix with real entries,

if λ in R is a root of the characteristic polynomial, how can i show that there exsists an eigenvector, x that is in R^n of A with eigenvalue λ.

is this also true if λ is not in R

Answer 1

Remember that the characteristic polynomial of A is det (A-λI), so if λ is a root of this polynomial in R, then det (A-λI) is zero, meaning ker (A-λI) ≠ ∅, so there is a nonzero vector x such that (A-λI)x = 0. By the distributive property, this means Ax - λIx = 0, so Ax - λx = 0 and Ax = λx, which means x is an eigenvector with eigenvalue λ.

This also works if λ is not in R, with the technical caveat that the vector you obtain will not actually be in R^n, but rather C^n (the set of n-tuples of complex numbers). This is a subtle conceptual distinction, and one which is not usually mentioned in introductory linear algebra classes, because for computational purposes, it really doesn't matter whether you work with real or complex numbers. But it is kind of important theoretically -- the idea of an eigenvector is one whose direction is not changed by the linear transformation, which seems to conflict with the fact that there are transformations on R^n which change the direction of every vector (example: any nontrivial 2×2 rotation matrix), yet these matrices still have eigenvalues and eigenvectors. The resolution is that the eigenvectors of the matrix are complex, so the fact that every vector in R² has its direction changed by a rotation does not conflict with the fact that there are vectors whose direction is not changed by such a matrix, because those vectors are not in R².

Answer 2

A little bit sloppy definition of eigenvector is:
Given an n-dimensional vectorspace over R and a regular matrix A then: a vector X (<>O, null vector) with property:
A(X) = lambda*X
is called an eigenvector of A and
lambda the corresponding eigenvalue.
The equation
det(A-lambda*E)= 0
is called the characteristic equation of A.
(E=identity-matrix)
Not all values of lambda are necessarily real, but since the vector space in your problem is R^n, we say that in that case there is no eigenvector.

If X is eigenvector of A then A(X)=l*X or
A(X)-l*X=O (null vector) or
A(X)-lE(X)= O or
(A-lE)X=O (null vector)
this means det(A-lE)=0 (null)

Answer 3

If x is a root of the characteristic polynomial,then det(A-xI)=0, so the matrix A-xI is singular. This means that there is a vector v with (A-xI)v=0. But then, Av=xv, so v is an eigenvector with eigenvalue x.