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Eigenvalues and eigenvectors

by Pseudo Epsilon
Tags: eigenvalues, eigenvectors
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Pseudo Epsilon
#1
May5-13, 01:18 AM
P: 102
can someone PLEASE explain eigenvalues and eigenvectors and how to calculate them or a link to a site that teaches it simply?
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Pseudo Epsilon
#2
May5-13, 01:18 AM
P: 102
Ive already read the wiki and asked my math teacher, he doesnt even know what they are.
UVW
#3
May5-13, 01:22 AM
P: 24
I think that Khan Academy does a great job explaining just that!

http://www.khanacademy.org/math/line...d-eigenvectors

Also, don't forget that there's a "Math & Science Learning Materials" forum on this website; it might be a better place to check in the future.

HallsofIvy
#4
May5-13, 08:04 AM
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PF Gold
P: 39,489
Eigenvalues and eigenvectors

Do you know what "vectors" and "linear transformations" are? Do you know what a "linear vector space" is?
HomogenousCow
#5
May5-13, 10:16 AM
P: 359
Quote Quote by Pseudo Epsilon View Post
Ive already read the wiki and asked my math teacher, he doesnt even know what they are.
That is sad to hear, eigenvectors and eigenvalues are very basic maths. Teachers are very underqualified these days.

A linear operator is a function that maps one vector space into another, there are certain vectors which when transformed by the linear operator, comes out as a scalar multiple of itself, the vector is the eigenvector and the multiple is the eigenvalue.
Pseudo Epsilon
#6
May5-13, 10:47 PM
P: 102
dont judge me but how does one map one vector space onto another?
Pseudo Epsilon
#7
May5-13, 10:59 PM
P: 102
he doesnt know what a vector space even is! And the wiki doesnt do much to even seperate it from vectors.
WannabeNewton
#8
May5-13, 11:23 PM
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Let ##V## be a vector space over ##F## and let ##T:V\rightarrow V## be a linear operator. We say ##v\in V\setminus \left \{ 0 \right \}## is an eigenvector of ##T## if there exists a ##\lambda\in F## such that ##T(v) = \lambda v##. We call ##\lambda## an eigenvalue of ##T##.

As an example, let ##V = M_{n\times n}(\mathbb{R})## and let ##T:V\rightarrow V,A \mapsto A^{T}##. We want to find the eigenvalues of ##T##. Let ##A\in V## such that ##T(A) = A^{T} = \lambda A##. Note that ##T(T(A)) = \lambda ^{2}A = (A^T)^T = A## hence ##A(\lambda^{2} - 1) = 0## and since eigenvectors have to be non-zero, this implies ##\lambda = \pm 1##.
Pseudo Epsilon
#9
May6-13, 08:37 AM
P: 102
thanks!


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