Crit. for a Mat. to be a Fund. Mat. to a Lin. Homo. Sys. of F. Order ODEs

Criterion for a Matrix to be a Fundamental Matrix to a Linear Homogeneous System of First Order ODEs

 Theorem 1: Let $\Phi$ be a solution to the matrix equation $X' = A(t)X$ on $J = (a, b)$. Then $\Phi$ is a fundamental matrix of the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$ on $J$ if and only if $\det \Phi (t) \neq 0$ for all $t \in J$.

Recall that if $Ax = b$ has a unique solution $x$ for every $b$ then $\det A \neq 0$. We use this property in the first direction of the proof below. We obtain a matrix equation $\Phi (\tau) \mathbf{a} = \xi$ and note that a unique solution $\mathbf{a}$ exists for every $\xi$, and so $\det \Phi (\tau) \neq 0$ and so $\det \Phi (t) \neq 0$ for all $t \in J$.

• Proof: Let $\Phi$ be a solution to the matrix equation $X' = A(t)X$ on $J = (a, b)$.
• $\Rightarrow$ Let $\Phi$ be a fundamental matrix of the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$. Then $\Phi$ is of the form:
(1)
\begin{align} \quad \Phi = \begin{bmatrix} \phi^{} & \phi^{} & \cdots & \phi^{[n]} \end{bmatrix} \end{align}
• Then by definition $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ is a linearly independent set of solutions to $\mathbf{x}' = A(t)\mathbf{x}$ and is hence a basis of the set of solutions $V$ for this system.
• Let $\xi \in \mathbb{R}^n$ and let $\phi$ be any solution to $\mathbf{x}' = A(t)\mathbf{x}$ with $\phi (\tau) = \xi$. Then for some $a_1, a_2, ..., a_n \in \mathbb{R}$ we have that:
(2)
\begin{align} \quad \phi(t) = a_1 \phi^{}(t) + a_2 \phi^{}(t) + ... + a_n\phi^{[n]} \end{align}
• Let $\mathbf{a} = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}$. Then the above equation can be rewritten as:
(3)
\begin{align} \quad \phi(t) = \Phi(t) \mathbf{a} \end{align}
• Plugging $t = \tau$ yields:
(4)
\begin{align} \quad \phi(\tau) &= \Phi (\tau) \mathbf{a} \\ \quad \xi &= \Phi (\tau) \mathbf{a} \end{align}
• $\Leftarrow$ Suppose that $\det \Phi (t) \neq 0$ for all $t \in J$. Since $\Phi$ is a solution to the matrix equation $X' = A(t)X$. Then the columns $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ is a set of solutions to $\mathbf{x}' = A(t)\mathbf{x}$. Moreover, since $\det \Phi (t) \neq 0$ for all $t \in J$ we have that $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ is a set of linearly independent solutions to $\mathbf{x}' = A(t) \mathbf{x}$ on $J$. Therefore $\Phi$ is a fundamental matrix to $\mathbf{x}' = A(t)\mathbf{x}$. $\blacksquare$