Why does finding eigenvalues of a system of differential equations always return the solution?

Question

I am currently learning about differential equations, and one method that we use to solve systems of homogenous systems of differential equations is to put the coefficients into a matrix, set the matrix equal to the zero vector, then find the eigenvalues and eigenvectors. I cannot find proof of this method anywhere and I've looked many times. This method is really stumping me :/

Answer 1

For $x'=Ax$ for a constant coefficient matrix $A$, we look for exponential solutions. It is basically just a guess, though it is motivated by what we already know about the scalar case. When we do that, we find that the exponential solutions are of the form $x_k(t)=e^{\lambda_k t} v_k$ where $v_k$ is an eigenvector with eigenvalue $\lambda_k$.

For a system of $n$ first order differential equations, you will have $n$ linearly independent solutions. This is essentially a consequence of the fact that the solution to the IVP is unique and is determined by the $n$ scalar parameters $x_1(t_0),\dots,x_n(t_0)$.

Now in linear algebra it turns out that "almost every" square matrix is diagonalizable (if we allow for complex eigenvalues), which means that it has $n$ linearly independent eigenvectors. In this case any solution can be represented as a combination of exponential solutions. So if you choose $A$ "at random" then you can solve $x'=Ax$ with exponential solutions only. Indeed in this case if $A=PDP^{-1}$ for a diagonal $D$ then $x(t)=e^{At} x(0)=Pe^{Dt} P^{-1} x(0)$, where $e^{Dt}$ is just the matrix with $e^{\lambda_k t}$ as its diagonal elements.

In general, however, not every matrix is diagonalizable. But still the formula $x(t)=e^{At} x(0)$ holds, so to handle the general case we just need a way to calculate the matrix exponential. It turns out that although not every square matrix is diagonalizable, every square matrix does have a Jordan normal form. And one can explicitly compute the matrix exponential of a nondiagonal Jordan normal form. For example, $\exp \left (\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} t \right ) = \begin{bmatrix} e^t & te^t \\ 0 & e^t \end{bmatrix}$. Thus one can at least in principle solve any system of the form $x'=Ax$ by using the eigenvectors and generalized eigenvectors as well.