# 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

Staff Emeritus
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!