3606 Fundamental Matrices

In the literature, solutions to linear systems often are expressed using square matrices rather than vectors. This is an elegant bookkeeping technique and a very compact, efficient way to express these formulas. As before, we state the definitions and results for a $2 \times 2$ system, but they generalize immediately to $n \times n$ systems.
We return to the system $$\boldsymbol{x}'=A(t)\boldsymbol{x}\tag{1}$$ with the general solution $$\boldsymbol{x}=c_1\boldsymbol{x}_1(t)+c_2\boldsymbol{x}(t)_2\tag{2}$$ where $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ are two independent solutions to $(1)$, and $c_1$ and $c_2$ are arbitrary constants. We form the matrix whose columns are the solutions $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$: $$\Phi(t)=\begin{pmatrix} \boldsymbol{x}_1\\\boldsymbol{x}_2 \end{pmatrix}=\begin{pmatrix} x_1&x_2\\y_1&y_2 \end{pmatrix}\tag{3}$$ Since the solutions are linearly independent, we called them a fundamental set of solutions, and therefore we call the matrix in $(3)$ a fundamental matrix for the system $(1)$.
Writing the general solution using $\Phi(t)$. As a first application of $\Phi(t)$, we can use it to write the general solution $(2)$ efficiently. For according to $(2)$, it is $$\boldsymbol{x}=c_1\begin{pmatrix} x_1\\y_1 \end{pmatrix}+c_2\begin{pmatrix} x_2\\y_2 \end{pmatrix}=\begin{pmatrix} x_1&x_2\\y_1&y_2 \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}$$ which becomes using the fundamental matrix $$\boldsymbol{x}=\Phi\boldsymbol{c}, \text{ where }\boldsymbol{c}=\begin{pmatrix} c_1\\c_2 \end{pmatrix}\tag{4}$$ Note that the vector $\boldsymbol{c}$ must be written on the right, even though the $c$'s are usually written on the left when they are the coefficients of the solutions $\boldsymbol{x}_i$.
Solving the IVP using $\Phi(t)$. We can now write down the solution to the IVP $$\boldsymbol{x'}=A(t)\boldsymbol{x}, \boldsymbol{x}(t_0)=\boldsymbol{x}_0\tag{5}$$ Starting from the general solution $(4)$, we have to choose the $\boldsymbol{c}$ so that the initial condition in $(6)$ is satisfied. Substituting $t_0$ into $(5)$ gives us the matrix equation for $\boldsymbol{c}$: $$\Phi(t_0)\boldsymbol{c}=\boldsymbol{x}_0$$ Since the determinant $|\Phi(t_0)|$ is the value at $t_0$ of the Wronskian of $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$, it is non-zero since the two solutions are linearly independent (Theorem 3 in the note on the Wronskian). Therefore the inverse matrix exists and the matrix equation above can be solved for $\boldsymbol{c}$: $$\boldsymbol{c}=\Phi(t_0)^{-1}\boldsymbol{x}_0$$ Using the above value of $\boldsymbol{c}$ in $(4)$, the solution to the IVP $(1)$ can now be written $$\boldsymbol{x}=\Phi(t)\Phi(t_0)^{-1}\boldsymbol{x}_0$$ Note that when the solution is written in this form, it's "obvious" that $\boldsymbol{x}(t_0) = \boldsymbol{x}_0$, i.e., that the initial condition in $(5)$ is satisfied.
An equation for fundamental matrices We have been saying "a" rather than "the" fundamental matrix since the system $(1)$ doesn't have a unique fundamental matrix: there are many ways to pick two independent solutions of $\boldsymbol{x'} = A\boldsymbol{x}$ to form the columns of $\Phi$. It is therefore useful to have a way of recognizing a fundamental matrix when you see one. The following theorem is good for this; we'll need it shortly.
Theorem 1 $\Phi(t)$ is a fundamental matrix for the system $(1)$ if its determinant $|\Phi(t)|$ is non-zero and it satisfies the matrix equation $$\Phi'=A\Phi\tag{7}$$ where $\Phi'$ means that each entry of $\Phi$ has been differentiated.
Proof. Since $|\Phi| \not\equiv 0$, its columns $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ are linearly independent, as we saw in the previous note. Let $\Phi=<script type="math/tex; mode=display">\begin{pmatrix}\boldsymbol{x}_1\\\boldsymbol{x}_2\end{pmatrix}$ . According to the rules for matrix multiplication $(7)$ becomes $$\begin{pmatrix} \boldsymbol{x}_1'\\\boldsymbol{x}_2' \end{pmatrix}=A\begin{pmatrix} \boldsymbol{x}_1\\\boldsymbol{x}_2 \end{pmatrix}=\begin{pmatrix} A\boldsymbol{x}_1\\A\boldsymbol{x}_2 \end{pmatrix}$$ which shows that $$\boldsymbol{x}_1'=A\boldsymbol{x}_1 \text{ and } \boldsymbol{x}_2'=A\boldsymbol{x}_2$$ this last line says that $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ are solutions to the system $(1)$.