18.03 Class 32, Apr 29

Real eigenvalues

Characteristic polynomial, eigenvalues, eigenvectors;
determinant, trace; Ray trajectories and solutions, normal modes.


Prologue on Linear Algebra.

Recall  [a b ; c d] [x ; y] = x[a ; c] + y[b ; d]

Write  A = [a b; c d]  and  alpha = [x ; y] .

I sketched the two vectors and this linear combination.

I ask: when is this product zero.

One way is for  x = 0 = y.  If  [a ; c]  and  [b ; d]  point in
different directions, this is the ONLY way. But if they lie along a
single line, we can find  x  and  y  so that the sum cancels.

We get a nonzero solution  [x ; y]  exactly when the slopes of the vectors
[a ; c]  and  [b ; d]  coincide:  c/a = d/b ,  or  ad - bc = 0.  This
combination of the entries in  A  is so important it's called the
"determinant" of the matrix:

det(A) = ad - bc

(It "determines" whether the matrix is invertible.) We have found:

A alpha = 0  has a nontrivial solution  alpha  exactly when  det(A) = 0.


I described a method for finding solutions to systems of equations.
As an example I took  x' = x + 2y,  y' = 2x + y.  The "coefficient matrix"
is  A = [1 2 ; 2 1] .  With  u = [x ; y]  the equation is

u' = Au.

I showed a Matlab/dfield plot of the vector field  v = Au.  I suggested
that it showed several solutions along straight rays from (or to) the origin.

For the vector  alpha  to be on such a trajectory you must have  v(alpha)
pointing in the same (or opposite) direction as  alpha.  This can be
expressed by saying that  v(alpha) = lambda alpha  for some number  lambda;
that is,

A alpha = lambda alpha ,   alpha  not  0.

(The symbol  lambda  is often used for this kind of thing; you may recall
it from Lagrange multipliers.)

Surprisingly enough, the first thing to do is to find the numbers  lambda
which make this possible. We want to do this using matrix algebra. There
is no matrix on the right hand side, but we can fix that using the
identity matrix

I = [1 0 ; 0 1]

This matrix has the property that  I alpha = alpha  for any vector  alpha.
So we can rewrite our equation as

A alpha = lambda I alpha

or

(A - lambda I) alpha = 0.

Now  lambda I = [lambda 0 ; 0 lambda] ,  and  A - lambda I  is  A  with
lambda  subtracted from the diagonal entries.  I also want  alpha  not  0,
and in the Prologue we found a criterion for the existence of such  alpha:

det(A - lambda I) = 0.

Let's work out what this is:

det[a-lambda b ; c d-lambda] = (a-lambda)(d-lambda) - bc

      = lambda^2 - (a+d) lambda + (ad-bc)

This is a degree 2 polynomial, the "characteristic polynomial"   p_A(lambda)
of  A.

The number  a+d ,  the sum of the diagonal entries of  A ,  is called the
"trace" of A,  tr(A) = a + d.  We already have a notation for the constant
term,  det(A):  so

p_A(lambda) = lambda^2 - (tr A) lambda + (det A)


In our example,  p_A(lambda) = lambda^2 - 2 lambda - 3 .

We are interested in the roots of the characteristic polynomial.
There are two; call them  lambda1  and  lambda2.  (The order doesn't matter.)
These are the "eigenvalues" of the matrix.

In our example, p_A(lambda) = (lambda + 1)(lambda - 3)  so the roots are

lambda1 = -1,  lambda2 = 3.

Let's next find the corresponding vectors:

For  lambda1  seek  alpha1  such that  A alpha1 = lambda1 alpha1
or equivalently   (A - lambda1 I) alpha1 = 0 .  Similarly for  lambda2.
These are the "eigenvectors" for the corresponding eigenvalues.

In our example,

A - lambda1 I = [2 2 ; 2 2]

and we are trying to find  alpha1 = [? ; ?]  not  0  such that

[2 2 ; 2 2] [? ; ?] = [0 ; 0].

Clearly   alpha1 = [1 ; -1]  or any nonzero multple will do.

Similarly,

A - lambda2 I = [-2 2 ; 2 -2] [? ; ?] = [0 ; 0]  has solution

alpha2 = [1 ; 1]  or any nonzero multiple.

I checked these on the Matlab picture.

So there are a few lines where the matrix acts simply by multiplication,
without changing directions.

We still have to solve the ODE. If  u(t)  is "ray solution," a solution
lying along a ray from (or to) the origin, then

u(t) = f(t) alpha.

We found the eigenvectors precisely because for this to work  alpha  must
be an eigenvector.  Say  alpha = alpha1.  Plug this into the equation:

u'(t) = f'(t) alpha1   versus

Au(t) = A f(t) alpha1 = f(t) A alpha1 = f(t) lambda1 alpha1.

Since  alpha1  is not zero, this forces

f'(t) = lambda1 f(t).

By focusing attention on ray solutions, we have gotten ourselves a single
ODE, and a very familiar one at that!

f(t) = c e^{lambda1 t}

For the moment just one solution will do, so take  c = 1:

u(t) = e^{lambda1 t} alpha1

is a ray solution, and

u(t) = e^{lambda2 t} alpha2

is another.

In our example, these are

u(t) = e^{-t} [1 ; -1]  ,   u(t) = e^{3t} [1 ; 1]

We have found solutions whose trajectories are rays from (or to) the origin.

As we saw, we can multiply these by constants to get other solutions.
More generally, because  u' = Au  is  LINEAR  and  HOMOGENEOUS,  any
linear combination of solutions is again a solution; and in fact

u = c1 e^{lambda1 t} alpha1 + c2 e^{lambda2 t} alpha2

is the GENERAL solution to  u' = Au.  (There is a minor warning here;
if  lambda1 = lambda2  you may not be able to find two eigenvectors pointing
in different directions, and then something else is called for. We'll
come back to this.)

In our example, the general solution is

u = c1 e^{-t} [1 ; -1] + c2 e^{3t} [1 ; 1]

We can solve for  c1  and  c2  using an initial condition:  say for example

u(0) = [1 ; 0].  Well,

u(0) = c1 [1 ; -1] + c2 [1 ; 1] = [c1+c2 ; -c1+c2]

and for this to be  [1 ; 0]  we must have  c1 = c2 = 1/2:

u(t) = (1/2)e^{-t} [1 ; -1] + (1/2)e^{3t} [1 ; 1] .

When  t  is very negative, -10, say, the first term is very big and the
second tiny: the solution is very near the line through [1 ; -1].
As  t  gets near zero, the two terms become comparable and the solution
curves around. As  t  gets large, 10, say, the second term is very big
and the first is tiny: the solution becomes asymptotic to the line through
[1 ; 1].

The general solution is a combination of the two normal modes.