Verifying a remark about fundamental matrices

There is a remark which gets used many times in a theorem I am trying to work through:

If $X(t)$ is a fundamental matrix of $x'(t) = A(t)x(t)$ (1), then $x(t) = X(t)X^{-1}(t_{0})x_{0}$.

In this setting, we are taking $x_{0} = x(t_{0})$.

To be honest, I'm not exactly sure what it is claiming exactly. Is just just a closed form of a particular solution to (1)? or is it a general solution to (1)? Since fundamental matrices can differ by multiples of constant matrices, this couldn't be describing all the solutions, right?

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From what book is this? – J. M. Oct 30 '11 at 0:47
We aren't using a book. It's just from my course notes. I'm referring to the first step in the proof of (a) of Theorem 3.1 on page 4. math.ualberta.ca/~mli/courses/current/524/chapter3.pdf – roo Oct 30 '11 at 1:34

The columns of a fundamental matrix span the space of solutions. In other words, if the columns of $X(t)$ are $x_1(t),\dots,x_n(t)$. Then $x(t)=c_1x_1(t)+\cdots+c_nx_n(t)$ is the general solution. We can write this in compact form by letting ${\bf c}=[c_1\;c_2\;\cdots\;c_n]^T$ and then $x(t)=X(t){\bf c}$.

Now suppose that you impose an initial condition on the solution: $x(t_0)=x_0$. Then $x_0=x(t_0)=X(t_0){\bf c}$. A theorem tells us that the determinant of a fundamental matrix is non-zero for all $t$. Therefore, $X(t_0)$ is an invertible matrix. Solving the system $X(t_0){\bf c}=x_0$ gives us ${\bf c}=X(t_0)^{-1}x_0$. Therefore, the solution is $x(t)=X(t)X(t_0)^{-1}x_0$.

To address the rest of your concerns. The fundamental existence/uniqueness theorem for linear systems tells us that there is exactly one solution to the initial value problem: $x'=Ax$, $x(t_0)=x_0$. So even though you can have different fundamental matrices, after the matrix has been multiplied by its inverse at $t_0$ and $x_0$, all of the differences cancel out and you get the unique solution.

Edit: To be clear (hopefully)... The system of equations has many different solutions and many different fundamental matrices, but an initial value problem only has one solution -- it's unique!

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That clarifies a lot. Thank you. – roo Oct 30 '11 at 1:36