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Describing Eigenvalues and Eigenvectors

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.


   A =  [  2   2 ]
        [ -1   5 ]

and let:

   X = [ x ]
       [ y ]

   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)


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


  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


   (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 |

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.


   M = P*D*P^(-1)
where P^(-1) is inverse matrix of P.


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

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

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


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


   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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