Wolfram Alpha:

```Eigenvectors and Eigenvalues
----------------------------

Eigenvectors are the directions along which a linear
transformation acts by stretching/compressing and/or
reversing a vector.  Eigenvalues give you the factors
by which the vectors are stretched/compressed.

The basic equation is:

A(αv) = αAv = αλv = λ(αv)

Where,

A is a matrix.
α is a scalar.
v is a vector

The Characteristic Polynomial
-----------------------------

The eigenvalues are found from:

(A - λI)x = 0

This has a solution only if det(A - λI) = 0

The eigenvectors are found from:

A - λI = 0

Example:

-    -
A = | 7  3 |
| 3 -1 |
-    -
-   -     -   -
Step 1.  λI = λ| 1 0 | = | λ 0 |
| 0 1 |   | 0 λ |
-   -     -   -

-    -     -   -
Step 2 :  A - λI = | 7  3 | - | λ 0 |
| 3 -1 |   | 0 λ |
-    -     -   -

-        -
= | 7-λ    3 |
| 3   -1-λ |
-        -

-        -
Step 3:  det| 7-λ    3 | = λ2 - 6λ - 16 = (λ - 8)(λ + 2) = 0
| 3   -1-λ |
-        -

∴ λ = 8 and -2

-     -
Step 4:  | -1  3 | = B
|  3 -9 |
-     -

For the eigenvectors (A - λI) = 0

-     -      -   -
| -7  3 | - 8| 1 0 | = 0
|  3 -1 |    | 0 1 |
-     -      -   -

Therefore,

-      -  - -     -         -
| -15  3 || x | = | -15x + 3y | = 0
|  3  -9 || y |   |   3x - 9y |
-      -  - -     -         -

=> x = 3 and y = 1

We can also see this another way:

Set Bv = 0

-     -  -  -     - -
| -1  3 || x1 | = | 0 |
|  3 -9 || x2 |   | 0 |
-     -  -  -     - -

-x1 + 3x2 = 0
3x1 - 9x2 = 0

∴ x1 = 3 and x2 = 1

-    -  - -      - -
| 7  3 || 3 | = 8| 3 |
| 3 -1 || 1 |    | 1 |
-    -  - -      - -

We follow the same procedure for λ = -2

-   -
| 9 3 | = B
| 3 1 |
-   -

x1 = 1 x2 = -3

-    -  -  -       -  -
| 7  3 ||  1 | = -2|  1 |
| 3 -1 || -3 |     | -3 |
-    -  -  -       -  -

Notes:  For an n x n matrix.

Tr = sum of eigenvalues.

det = product of eigenvalues.

Significance
------------

There are many problems that can be modeled with linear
transformations, and the eigenvectors give very simple
solutions. For example, consider the system of linear
differential equations

dx/dt = ax + by

dy/dt = cx + dy

Solving this system directly is complicated because the
2 equations are coupled to each other.

-     -     -   -  - -
| dy1/dt | = | a b || y1 |
| dy2/dt |   | c d || y2 |
-     -     -   -  - -
.
y = Ay

Solutions are of the form:

y1 = αexp(λt)

and,

y2 = βexp(λt)

We can write this as:
- -
y = exp(λt)| α |
| β |
- -

= exp(λt)e

d(αexp(λt))/dt = λexp(λt)e

λexp(λt)e = λexp(λt)e

Now Ae = λe

λexp(λt)e = Aexp(λt)e

λe = Ae

This is a solution for one eigenvalue.  We can use the same
argument to get a solution for thesecond eigenvalue.  The
general solution is obtained by taking linear combinations
of these two solutions, so the general solution is of the
form:

y = c1exp(λ1)e1 + c2exp(λ2)e2

-  -             -  -             -  -
| y1 | = c1exp(λ1)| α1 | + c2exp(λ2)| α2 |
| y2 |            | β1 |            | β2 |
-  -             -  -             -  -

Where c1 and c2 are constants determined using the initial
conditions.

Example:

dx/dt = 4x - y

dy/dt = 2x + y

We can write this in matrix form as:

-     -     -    -  - -
| dx/dt | = | 4 -1 || x |
| dy/dt |   | 2  1 || y |
-     -     -    -  - -

Using the above processes the eigenvalues and eigenvectors
are:
- -
λ1 = 3  e1 = | 1 |
| 1 |
- -

-   -
λ2 = 2  e1 = | 1/2 |
|  1  |
-   -

- -
| x | = Cexp(λ1t)e1 + Dexp(λ2t)e2
| y |
- -

- -             - -             -   -
| x | = Cexp(3t)| 1 | + Dexp(2t)| 1/2 |
| y |           | 1 |           |  1  |
- -             - -             -   -

x = Cexp(3t) + Dexp(2t)/2

y = 2Cexp(3t) + Dexp(2t)

Geometry of Real and Complex Eigenvalues
----------------------------------------

When the eigenvalues are real the vectors along the
eigenvectors (bases) are stretched, compressed or
reversed.

/
/ λx
x //
//

Aξ = λξ

When the eigenvalues are complex the story is a
different.  Lets go through the analysis to see
what happens.

If λ = a + ib is complex then the eigenvector,
ξ = x + iy, is complex.

The eigenvalue equation can now be written as:

A(x + iy) = (a + ib)((x + iy)

THerefore,

Ax + iAy = (ax - by) + i(bx + ay)

Which can be broken down as follows:

Ax = (ax - by)

and,

Ay = (bx + ay)

Now lets change the basis of the vector space:

e1 = x = Re(ξ)

e2 = y = Im(ξ)

-     -  -  -
[A]e = | α1 -α2 || e1 |
| α2  α1 || e2 |
-     -  -  -

-          -
= | α1e1 - α2e2 |
| α2e1 + α1e2 |
-          -

Lets look at amore complicated example using the
matrix, J1 from so (3,1) that generates rotations
in the yz about the x axis.

-        -
| 0 0  0 0 |
iJ1 = | 0 0  0 0 |
| 0 0  0 1 |
| 0 0 -1 0 |
-        -

The eigenvalues, λ, are - 0, -i, i

The eigenvectors, ξ, are:

- -    - -    - -    -  -
| 1 |  | 0 |  | 0 |  |  0 |
| 0 |, | 1 |, | 0 |, |  0 |
| 0 |  | 0 |  | i |  | -i |
| 0 |  | 0 |  | 1 |  |  1 |
- -    - -    - -    -  -

-        -  - -     - -      - -
| 0 0  0 0 || 1 |   | 0 |    | 1 |
| 0 0  0 0 || 0 | = | 0 | = 0| 0 |
| 0 0  0 1 || 0 |   | 0 |    | 0 |
| 0 0 -1 0 || 0 |   | 0 |    | 0 |
-        -  - -     - -      - -

-        -  - -     - -      -  -
| 0 0  0 0 || 0 |   | 0 |    |  0 |
| 0 0  0 0 || 1 | = | 0 | = 0|  1 |
| 0 0  0 1 || 0 |   | 0 |    |  0 |
| 0 0 -1 0 || 0 |   | 0 |    |  0 |
-        -  - -     - -      -  -

-        -  - -     -  -       -  -
| 0 0  0 0 || 0 |   |  0 |     |  0 |
| 0 0  0 0 || 0 | = |  0 | = -i|  0 |
| 0 0  0 1 || i |   |  1 |     |  i |
| 0 0 -1 0 || 1 |   | -i |     |  1 |
-        -  - -     -  -       -  -

-        -  - -     -  -       -  -
| 0 0  0 0 ||  0 |   | 0 |     |  0 |
| 0 0  0 0 ||  0 | = | 0 | =  i|  0 |
| 0 0  0 1 || -i |   | 1 |     | -i |
| 0 0 -1 0 ||  1 |   | i |     |  1 |
-        -  - -     -  -       -  -

Look at the eigenvector corresponding to i.

-  -     - -      -  -
|  0 |   | 0 |    |  0 |
ξ = |  0 | ≡ | 0 | + i|  0 |
| -i |   | 0 |    | -1 |
|  1 |   | 1 |    |  0 |
-  -     - -      -  -

= Re(ξ) + iIm(ξ)

= e1 + ie2

Where we have defined e1 and e2 as our basis vectors.

Therefore,

- -            -  -
| 0 |          |  0 |
e1 = | 0 | and e2 = |  0 |
| 0 |          | -1 |
| 1 |          |  0 |
- -            -  -

Thus A represents a rotation in the yz plane.  We would
also see this if we did a similar calculation for λ = -i.

Quantum Mechanics
-----------------

The possible states of a quantum mechanical system are
represented by unit vectors (called state vectors)
residing in a complex Hilbert space (state space).
In any measurement on a quantum mechanical system, the
value of the observable attained will be one of the
eigenvalues of the Hermitian operator that corresponds
to the observable.  The eigenvalues of each eigenstate
correspond to the allowable values of the quantity
being measured.  Following the measurement, the state
of the system will be the corresponding eigenvector.
The eigenvectors of a Hermitian operator form a complete
set.  That is they from an orthonormal basis.  The
eigenvalues must be real.

An eigenvalue which corresponds to two or more different
linearly independent eigenvectors is said to be DEGENERATE,

-     -
| 0 1 1 |
| 1 0 1 |
| 1 1 0 |
-     -

λ1 = -1   λ2 = -1  λ3 = 2

- -    -  -    - -
| 1 |  | -1 |  | 1 |
| 1 |  |  0 |  | 1 |
| 0 |  |  1 |  | 1 |
- -    -  -    - -
```