 In a kind and gentle universe every matrix will have a complete set of eigenvalues and every one of these will be real We don't live in that universe Where were we? Oh, yeah If we can write a first-order linear system of differential equations in the form dx equals ax Then our theorem says that if lambda v is an eigenvalue eigenvector pair for a C e to power lambda t v is a solution to the differential equation By linearity any linear combination of these solutions is also a solution You should probably prove that Now if all the eigenvalues of a are real and distinct we can find the general solution this way, but what if they're not? So one possibility is our eigenvalues might be complex numbers To handle this situation. We'll go back to 1722 when the English mathematician ever Hamden-Rov described the following relationship If a and b are real numbers then e to power a plus b i can be split into a real component e to the a times a trigonometric component cosine b plus i sine b and This suggests that if lambda equals a plus bi x is an eigenvalue eigenvector pair for a then C e to power a t cosine b t plus i sine b t x will be a solution to our system of differential equations Except there's a problem if our original system of differential equations involved real Coefficients only we would expect our solution to involve real coefficients only and this isn't real So what can we do? In order to reduce this complex solution to a real solution. We need the following theorems First suppose a is a matrix with real entries and lambda v are an eigenvalue eigenvector pair That the conjugates of lambda and v are also an eigenvalue eigenvector pair This is easy to prove so you should actually prove it It's somewhat more difficult to prove another important theorem Let dx equal a x be a linear system of differential equations if y equals f of x plus i G of x is a solution where f of x and g of x are real-valued then f of x and g of x separately are also solutions and This isn't too difficult to prove though. We do have to establish that imaginary numbers act just like real numbers when it comes to differentiation However, once we've established that we can handle complex solutions of real differential equations So for example suppose I have this linear system We can rewrite this in operator form the coefficient matrix will have characteristic polynomial And so we'll have eigenvalues plus or minus 2i For lambda equals 2i the eigenvector will be and so an eigenvector will have x2 equals 2i And x1 equals minus 1 This gives us the solution e to power 2i t times the vector minus 1 to i Now we do need to split this into the real and complex portions of the functions So we'll use them off theorem e to power 2i t times our vector is going to be our vector times Cosine 2t plus i sine 2t We'll rewrite our vector in terms of its real and complex components multiply out all of our terms and Simplify and so now we've written our solution e to power 2i t times a vector in terms of a sum of a real part Which we can write as a single vector Plus a complex part Which we can write as i times a single vector And our theorem says that the real part forms a solution our complex part forms a solution and Linearity guarantees that any linear combination of these two will also form a solution And so we have the general solution Or do we Remember that this was just for the eigenvalue lambda equals 2i There was another eigenvalue lambda equals minus 2i and that should give us another solution Well, let's be clever about it While we could find the eigenvector for the eigenvalue minus 2i This will just be the conjugate of the eigenvector for 2i and Consequently this eigenvalue eigenvector pair will give us a solution the conjugate of the eigenvector Times the conjugate of the solution But that will just be the conjugate of the eigenvector times the solution and we already determined what that is and Since we've now split this into a real and complex part we can find the conjugate directly We'll just change the sign of the complex part and again the real and complex parts give us solutions The real part here is the same and so it'll produce the same type of term in the general solution and While the complex part looks different remember that this is going to be multiplied by some arbitrary Constant and we could absorb that minus and that gives us the same solution for the complex part