Cha. of Bas. Thms. for Fund. Mat. of a Lin. Homo. Sys. of F. O. ODEs

# The Change of Basis Theorems for Fundamental Matrices of a Linear Homogeneous System of First Order ODEs

Theorem 1: If $\Phi$ is a fundamental matrix to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$ on $J = (a, b)$ then for any nonsingular constant $n \times n$ matrix $C$, $\Phi C$ is also a fundamental matrix to $\mathbf{x}' = A(t)\mathbf{x}$ on $J$. |

**Proof:**Let $\Phi$ be a fundamental matrix to $\mathbf{x}' = A(t)\mathbf{x}$ and let $C$ be a nonsingular constant matrix. Since $\Phi$ is a fundamental matrix, $\Phi$ is a solution to the matrix equation $X' = A(t) X$, that is, $\Phi' = A(t) \Phi$. So:

\begin{align} \quad (\Phi C)' &= \Phi' C \\ &= [A(t) \Phi] C \\ &= A(t) (\Phi C) \end{align}

- Therefore $\Phi C$ is a solution to the matrix equation $X' = A(t) X$. Furthermore, since $C$ is a nonsingular constant matrix, $\det C \neq 0$, and since $\Phi$ is a fundamental matrix to $\mathbf{x}' = A(t)\mathbf{x}$ on $J$ we have that $\det \Phi (t) \neq 0$ on $J$.

- Therefore, for all $t \in J$

\begin{align} \quad \det (\Phi (t) C) = \det (\Phi (t)) \det (C) \neq 0 \end{align}

- Since $\Phi C$ is a solution to the matrix equation $X' = A(t) X$ and $\det (\Phi (t) C) \neq 0$ for all $t \in J$, from the result proven on the Criterion for a Matrix to be a Fundamental Matrix to a Linear Homogeneous System of First Order ODEs page we have that $\Phi C$ is a fundamental matrix to $\mathbf{x}' = A(t)\mathbf{x}$ on $J$. $\blacksquare$

Theorem 2: If $\Phi$ and $\Psi$ are fundamental matrices to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$ on $J = (a, b)$ then there exists a nonsingular constant matrix $C$ such that $\Phi C = \Psi$. |

**Proof:**Let $\Phi$ and $\Psi$ be fundamental matrices to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$ on $J$. Since $\Phi$ is a fundamental matrix, $\det \Phi \neq 0$ on all $J$ and so $\Phi^{-1}$ exists on all of $J$. By the product rule we have that:

\begin{align} \quad 0 = \frac{d}{dt} I = \frac{d}{dt} (\Phi \Phi^{-1}) = \Phi' \Phi^{-1} + \Phi (\Phi^{-1})' \end{align}

- Therefore:

\begin{align} \quad \Phi (\Phi^{-1}) = -\Phi' \Phi^{-1} \\ \quad \Phi^{-1} = -\Phi^{-1} \Phi' \Phi^{-1} \quad (*) \end{align}

- Now by the product rule, and by using $(*)$, $\Phi' = A(t)\Phi$, and $\Psi' = A(t)\Psi$ we have that:

\begin{align} \quad (\Phi^{-1} \Psi)' &= (\Phi^{-1})' \Psi + \Phi^{-1} \Psi' \\ & \overset{(*)} = [-\Phi^{-1} \Phi' \Phi^{-1}] \Psi + \Phi^{-1} \Psi' \\ &= -\Phi^{-1} A(t)\Phi \Phi^{-1} \Psi + \Phi^{-1} A(t) \Psi \\ &= -\Phi^{-1} A(t) \Psi + \Phi^{-1} A(t) \Psi \\ &= 0 \end{align}

- So there exists a constant matrix $C$ such that $\Phi^{-1} \Psi = C$. So $\Phi C = \Psi$. Furthermore, $\det C \neq 0$ since $\det \Phi \neq 0$ and $\det \Psi \neq 0$ on $J$, so $C$ is also nonsingular. $\blacksquare$