Homogeneous Linear System with Constant Coefficients

So in this post I am going to talk about solving homogeneous linear systems with constant coefficients.  It will be assumed that anyone reading this knows basic vector and matrix operations.  We are going to solve..

As with normal, non system differential equations, the guess for the solution will be an exponential function.

Notice that we need to multiply the exponential by the vector u.  The lambda in the exponential is our old friend the eigenvalue, and the corresponding vector is called the eigenvector.  We MUST multiply the exponential by the eigenvector, otherwise it wouldn't make sense to have a scalar solution to a vector problem.  All we need to do now is plug this into the system.

It is tempting to try and divide by the vector "u" but this CANNOT be done! Division doesn't exist in vectors, it isn't defined!  Notice that we have multiplied the eigenvalue by the identity matrix, this is analogous to multiplying by 1 so we aren't changing anything.  The only reason for doing this was to be able to combine the matrix A with the eigenvalues.  In order to solve this system we first realize that the vector x is not equal to the zero vector, yet the product of x and A-lambda I is equal to zero.  This implies that the matrix A-lambda I is singular which means its determinant is equal to zero.  Therefore the problem is now..

The resulting polynomial is termed the characteristic equation of the matrix.

Solve: (I am just going to show the math with little explanation, as I use this blog as a study tool.)

So what happens if the eigenvalues are complex?  Our old friend Euler's Formula comes into play, just as it did in normal differential equations.  (By normal, I mean non-system.)

So in this next derivation, we will assume the eigenvalues to our system are complex.  Complex numbers always come in pairs remember!  The beauty of this is, when you go to find the eigenvectors, you only need to use one of the complex eigenvalues, the other eigenvector you will get using the other one will just be the conjugate of the first eigenvector.

It can be shown that taking the real part of the imaginary part of this expression (in other words, using the second half without the "i") will also be a solution.  We only used one eigenvector but we would have the same results if we used the other one.  Therefore the two solutions are..

Solve:

Solve:

Notice that we wrote the answer as one matrix called the fundamental matrix.