The Det. of a Fund. Mat. to a Lin. Homo. System of First Order ODEs

The Determinant of a Fundamental Matrix to a Linear Homogeneous System of First Order ODEs

Theorem 1: If $\Phi$ is a solution to the matrix equation $X' = A(t) X$ on an interval $J = (a, b)$ and if $\tau \in J$ then for all $t \in J$, $\displaystyle{\det \Phi (t) = \det \Phi (\tau) \cdot \mathrm{exp} \left ( \int_{\tau}^{t} \mathrm{tr} (A(s)) \: ds \right )}$.

• Proof: Let:

(1)

\begin{align} \quad \Phi(t) = \begin{bmatrix} \phi_1^{[1]}(t) & \phi_1^{[2]}(t) & \cdots & \phi_1^{[n]}(t) \\ \phi_2^{[1]}(t) & \phi_2^{[2]}(t) & \cdots & \phi_2^{[n]}(t) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t) & \phi_n^{[2]}(t) & \cdots & \phi_n^{[n]}(t) \\ \end{bmatrix} \end{align}

• We take the derivative of the determinant of the matrix $\Phi$ with respect to $t$ to get::

(2)

\begin{align} \quad \frac{d}{dt} \det \Phi(t) &= \lim_{h \to 0} \frac{ \det \Phi (t + h) - \det \Phi (t)}{h} \\ &= \lim_{h \to 0} \frac{1}{h} \left ( \begin{vmatrix} \phi_1^{[1]}(t + h) & \phi_1^{[2]}(t + h) & \cdots & \phi_1^{[n]}(t + h) \\ \phi_2^{[1]}(t + h) & \phi_2^{[2]}(t + h) & \cdots & \phi_2^{[n]}(t + h) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t + h) & \phi_n^{[2]}(t + h) & \cdots & \phi_n^{[n]}(t + h) \\ \end{vmatrix} - \begin{vmatrix} \phi_1^{[1]}(t) & \phi_1^{[2]}(t) & \cdots & \phi_1^{[n]}(t) \\ \phi_2^{[1]}(t) & \phi_2^{[2]}(t) & \cdots & \phi_2^{[n]}(t) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t) & \phi_n^{[2]}(t) & \cdots & \phi_n^{[n]}(t) \\ \end{vmatrix} \right ) \end{align}

• We add/subtract certain matrices as follows:

(3)

\begin{align} \quad \Phi'(t) &= \lim_{h \to 0} \frac{1}{h} \left ( \begin{vmatrix} \phi_1^{[1]}(t + h) & \phi_1^{[2]}(t + h) & \cdots & \phi_1^{[n]}(t + h) \\ \phi_2^{[1]}(t + h) & \phi_2^{[2]}(t + h) & \cdots & \phi_2^{[n]}(t + h) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t + h) & \phi_n^{[2]}(t + h) & \cdots & \phi_n^{[n]}(t + h) \\ \end{vmatrix} - \begin{vmatrix} \phi_1^{[1]}(t) & \phi_1^{[2]}(t) & \cdots & \phi_1^{[n]}(t) \\ \phi_2^{[1]}(t + h) & \phi_2^{[2]}(t + h) & \cdots & \phi_2^{[n]}(t + h) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t + h) & \phi_n^{[2]}(t + h) & \cdots & \phi_n^{[n]}(t + h) \\ \end{vmatrix} \\ + \begin{vmatrix} \phi_1^{[1]}(t) & \phi_1^{[2]}(t) & \cdots & \phi_1^{[n]}(t) \\ \phi_2^{[1]}(t + h) & \phi_2^{[2]}(t + h) & \cdots & \phi_2^{[n]}(t + h) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t + h) & \phi_n^{[2]}(t + h) & \cdots & \phi_n^{[n]}(t + h) \\ \end{vmatrix} - \begin{vmatrix} \phi_1^{[1]}(t) & \phi_1^{[2]}(t) & \cdots & \phi_1^{[n]}(t) \\ \phi_2^{[1]}(t) & \phi_2^{[2]}(t) & \cdots & \phi_2^{[n]}(t) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]}(t) & \phi_n^{[2]}(t) & \cdots & \phi_n^{[n]}(t) \\ \end{vmatrix} \right ) & \vdots \\ & = \left ( \begin{vmatrix} \phi_1^{[1]'} & \phi_1^{[2]'} & \cdots & \phi_1^{[n]'} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \end{vmatrix} + \begin{vmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ \phi_2^{[1]'} & \phi_2^{[2]'} & \cdots & \phi_2^{[n]'} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \end{vmatrix} + \cdots + \begin{vmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]'} & \phi_n^{[2]'} & \cdots & \phi_n^{[n]'} \end{vmatrix} \right ) \end{align}

• Now notice that since $\Phi$ is a solution to the matrix equation $X' = A(t)X$ we have that $\Phi' = A(t)\Phi$. So for each $i, j \in \{ 1, 2, ..., n \}$ we have that:

(4)

\begin{align} \quad \phi_i^{[j]'} = \sum_{k=1}^{n} a_{i,k} \phi_{k}^{[j]} \end{align}

• Using this substitution above yields:

(5)

\begin{align} \frac{d}{dt} \det \Phi(t) = \left ( \begin{vmatrix} \sum_{k=1}^{n} a_{1,k} \phi_{k}^{[1]} & \sum_{k=1}^{n} a_{1,k} \phi_{k}^{[2]} & \cdots & \sum_{k=1}^{n} a_{1,k} \phi_k^{[n]} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \end{vmatrix} + \begin{vmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ \sum_{k=1}^{n} a_{2,k} \phi_{k}^{[1]} & \sum_{k=1}^{n} a_{2,k} \phi_{k}^{[2]} & \cdots & \sum_{k=1}^{n} a_{2,k} \phi_k^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \end{vmatrix} + \\ \cdots + \begin{vmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \sum_{k=1}^{n} a_{n,k} \phi_{k}^{[1]} & \sum_{k=1}^{n} a_{n,k} \phi_{k}^{[2]} & \cdots & \sum_{k=1}^{n} a_{n,k} \phi_k^{[n]} \\ \end{vmatrix} \right ) \end{align}

• For each $k \in \{ 1, 2, ..., n \}$ we can write an equivalent form of the determinant for each of $k$-determinants in the equation above by using elementary row operations to get:

(6)

\begin{align} \quad \frac{d}{dt} \det \Phi(t) &= \left ( \begin{vmatrix} a_{1,1} \phi_1^{[1]} & a_{1,1} \phi_1^{[2]} & \cdots & a_{1,1} \phi_1^{[n]} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \end{vmatrix} + \begin{vmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ a_{2,2} \phi_2^{[1]} & a_{2,2}\phi_2^{[2]} & \cdots & a_{2,2} \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \end{vmatrix} + \\ \cdots + \begin{vmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,n} \phi_n^{[1]} & a_{n,n} \phi_n^{[2]} & \cdots & a_{n,n} \phi_n^{[n]} \\ \end{vmatrix} \right ) \\ &= a_{1,1} \det \Phi (t) + a_{2,2} \det \Phi (t) + ... + a_{n,n} \det \Phi (t) \\ &= \det \Phi(t) \sum_{k=1}^{n} a_{k,k} \\ &= \det \Phi(t) \cdot \mathrm{tr} (A(t)) \end{align}

• Therefore $\frac{d}{dt} \det \Phi(t) = \det \Phi(t) \cdot \mathrm{tr} (A(t))$. For $\tau \in J$ we can use integrating factors to get that for all $t \in J$:

(7)

\begin{align} \quad \det \Phi (t) = \det \Phi (\tau) \cdot \mathrm{exp} \left ( \int_{\tau}^{t} \mathrm{tr} (A(s)) \: ds \right ) \quad \blacksquare \end{align}