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### Describing Eigenvalues and Eigenvectors

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Date: 04/05/98 at 16:10:40
From: Charlie Higgins
Subject: Eigenvalues

After working 30 years in industry, I went through an alternative
Teacher Certification program and got a teaching certificate. I now
teach at a high school. I do not have a degree in mathematics and my
degree is in Chemistry, but I do have a minor in math with enough
hours to qualify for the math certificate.

The new Algebra II books have a chapter on matrices (basic). One of
my students asked me the other day "What is an Eigenvalue?" I did
not take linear algebra in college and could not answer him. Could
someone explain to me in terms where I can explain it to my student
what an Eigenvalue is?

Warmest Regards,

Charlie Higgins
```

```
Date: 04/06/98 at 08:13:58
From: Doctor Jerry
Subject: Re: Eigenvalues

Hi Charlie,

If A is an n by n matrix, then a real or complex number w is an
eigenvalue of A if there is a nonzero n by 1 matrix Y for which
A*Y = w*Y. The vector Y is an eigenvector.

If you think about 2 by 1 matrices X as like vectors in an (x,y)-plane
and A is a given matrix, then multiplication by A is like a
deformation of the plane. The point with position vector X is
transformed to AX. If A is a rotation matrix, for example, which
rotates each point X about the origin, then original position is X
and, after rotation, AX. Or A can be thought about as an elastic
deformation of the plane, where the plane is an elastic sheet. X is
the position vector of a point before deformation and AX is the
position vector after deformation.

Let:

A =  [  2   2 ]
[ -1   5 ]

and let:

X = [ x ]
[ y ]

Then:

AX =  [ 2x+2y ]
[ -x+5y ]

One can ask: under this elastic deformation, is there a direction
along which there is no rotation, just pure stretching? So, one wants
a direction X for which:

AX = wX

where w is a scalar. wX would be pure stretching, since the direction
wouldn't change.

So, we want X and w so that (A-wI)X = 0, where I is 2 by 2 identity.
This system of equations has a nontrivial solution only if
det(A-wI) = 0.

A-wI = [ 2-w   2  ]
[ -1   5-w ]

det(A-wI) = (2-w)(5-w)+2 = (w-3)(w-4)

So, eigenvalues  are 3 and 4.

This part of linear algebra is very important to engineers,
physicists, mathematicians, statisticians, and others.

-Doctor Jerry,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 04/06/98 at 08:33:43
From: Doctor Anthony
Subject: Re: Eigenvalues

An easy way to visualize what eigenvalues and eigenvectors are is to
think in 2 dimensions and consider a matrix transformation in
the plane.

A matrix M,  say:

M = [a   b]
[c   d]

operating on a vector u will in general change u by enlarging,
rotating, reflecting, or shearing u to some new vector v. Now suppose
we want to look at the SPECIAL vector u which apart from enlargement
or contraction remains in same direction as u after the transformation
by the matrix M. What we require is:

M*u =  k*u  where k is some scalar factor (the enlargement factor)

So:

[a   b][x] = k[x]
[c   d][y]    [y]

and

ax + by = kx
cx + dy = ky

(a-k)x +     by =  0
cx + (d-k)y =  0

Now we have two homogeneous equations (no constant terms only terms in
x and y) with two unknowns, and we can only have solutions other than
x = 0, y = 0 if:

a-k      b
---- = ----
c      d-k

then:

(a-k)(d-k) - bc = 0               (Equation 1)

This condition is usually given in the form:

determinant |a-k     b|  =  0
|  c   d-k|

If we multiply out equation (1) we get:

ad - k(a+d) + k^2 - bc = 0

k^2 - k(a+d) + ad - bc = 0

This is a quadratic in k of the form:

k^2 - k(Trace M) + Det M = 0

(Trace = sum of leading diagonal of M, Det M = determinant of M)

The two values of k which are solutions to this equation are the
EIGENVALUES of M. Suppose we call them k1 and k2; then we can also
find the unique vectors u1 and u2 that remain unchanged (apart from
enlargement) when matrix M is used to transform them.

We have:

(a-k1)x + by = 0

(a-k1)x = - by

x     -b
--- = ----
y    a-k1

and we get:

u1 = |x|  =  |  b |
|y|     |k1-a|

and so associated with the eigenvalue k1 we get the eigenvector u1
given by:

u1 = |  b |
|k1-a|

and there will be another eigenvector u2 associated with the
eigenvalue k2.

If we express any other vector in terms of u1 and u2 as unit base
vectors, then the transformation represented by M can be given much
more simply as the matrix

[k1    0]   =  D
[ 0   k2]

This has many applications, and one that you might meet fairly soon is
finding powers of matrices. You 'diagonalize' a matrix M by expressing
it in terms of u1, u2, and the above matrix, D, with leading diagonal
k1 and k2.

If P = matrix made up of u1 and u2 as columns of P then:

M*P = P*D

which follows from M*u1 = k1*u1 and  M*u2 = k2*u2.

So:

M = P*D*P^(-1)

where P^(-1) is inverse matrix of P.

Then:

M^2 = P*D*P^(-1)*P*D*P^(-1)

= P*D*D*P^(-1)

= P*D^2*P^(-1)

Similarly:

M^3 = P*D^3*P^(-1)

and:

M^n = P*D^n*P^(-1)

and of course D^n is simply:

[k1^n     0]
[ 0    k2^n]

I have just touched on the subject, but you will find that
understanding this basic work on eigenvalues and eigenvectors will
make the work of transformation geometry a lot clearer than it might
otherwise be.

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Algebra

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