Eigenvalues and eigenvectors

1. May 5, 2013

Pseudo Epsilon

can someone PLEASE explain eigenvalues and eigenvectors and how to calculate them or a link to a site that teaches it simply?

2. May 5, 2013

Pseudo Epsilon

Ive already read the wiki and asked my math teacher, he doesnt even know what they are.

Last edited by a moderator: May 5, 2013
3. May 5, 2013

4. May 5, 2013

HallsofIvy

Do you know what "vectors" and "linear transformations" are? Do you know what a "linear vector space" is?

5. May 5, 2013

HomogenousCow

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.

6. May 5, 2013

Pseudo Epsilon

dont judge me but how does one map one vector space onto another?

7. May 5, 2013

Pseudo Epsilon

he doesnt know what a vector space even is! And the wiki doesnt do much to even seperate it from vectors.

8. May 5, 2013

WannabeNewton

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 $\lambda = \pm 1$.

9. May 6, 2013

thanks!