In this video, we start our investigation
of linear homogeneous equations. To do so,
we define an object called an eigenvalue.
So, we’re looking at matrix equations that
look like this. This matrix will now be assumed
to be constant.
If we had a simple linear equation like this,
we could solve it easily; the solution is
exponential. Using this as our guide, we might
expect to solve this like so. This does work,
but the issue is that defining this exponential
is very nontrivial; we do not simply put each
entry of the matrix into an exponent. For
now, we’ll try something similar, but simpler.
We can define e to the lambda t for a number
lambda, so let’s try to come close to this
by looking for a solution of the form ce^(lambda
t)
Can this be a solution? Lets see. We want
it to satisfy this. The derivative is lambda
c e^(lambda t), and we set it equal to A ce^(lambda
t)
Because e^(lambda t) is never 0, we can divide
both sides of the equation by it:
Lambda c = Ac
This motivates the following definition.
If A is a square matrix, an eigenvalue of
A is a number lambda such that Av = lambda
v for some nonzero vector v. The vector v
is called an eigenvector associated with the
eigenvalue lambda
In this video, we have defined eigenvalues
and eigenvectors. We will shortly use these
to solve systems of linear differential equations.
