Liouville's formula
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Problem. Prove Liouville's formula \(W=W_0 e^{\int \mathrm{tr}\ A dt}\) for the Wronskian determinant of the linear system \(\dot{\mathbf{x}} = A(t)\mathbf{x}\).
Proof. Let \(\mathbf{x}(t)\in\mathbb{R}^n\) and \(A(t)\in\mathbb{R}^{n\times n}\). The solution to the system \(\dot{\mathbf{x}} = A(t)\mathbf{x}\) is in the following form.
\[\mathbf{x}(t) = e^{\int_0^t A(s)\ ds}\mathbf{x}(0)\]Since \(e^{\int_0^t A(s)\ ds}\) is invertible, there would have \(n\) linearly independent solutions \(\mathbf{x}_1,\dots,\mathbf{x}_n\). Hence, we can write the Wronskian in the following form.
\[W(t) = \begin{bmatrix} \mathbf{x}_1(t) & \cdots & \mathbf{x}_n(t) \end{bmatrix} = e^{\int_0^t A(s)\ ds} \begin{bmatrix} \mathbf{x}_1(0) & \cdots & \mathbf{x}_n(0) \end{bmatrix} = e^{\int_0^t A(s)\ ds} W(0)\]Therefore \(\det W(t) = \det e^{\int_0^t A(s)\ ds}\det W(0)\).
Claim. \(\det e^A = e^{\mathrm{tr}\ A}\)
Proof of Claim. Decompose \(A=UTU^{-1}\), where \(T\) is an upper triangular matrix. Then we can show that \(\det e^{B}=e^{T}\) by the following derivation.
\[\begin{aligned} e^{A} &= \sum_n \frac{1}{n!} A^n = \sum_n\frac{1}{n!} (UTU^{-1})^n = \sum_n\frac{1}{n!} UT^n U^{-1}\\ &= U \left(\sum_n\frac{1}{n!}\right) U^{-1} = U e^T U^{-1}\\ \Rightarrow \det e^A &= \det U \det e^T \det U^{-1} = \det e^T \end{aligned}\]Since \(T\) is upper-triangular, then \(e^T\) is also upper-triangular. Due to the property of upper-triangular matrices where \([e^T]_{ii} = e^{T_{ii}}\), and \(\det T = \prod_i T_ {ii}\), we can prove that \(\det e^T = e^{\mathrm{tr}\ T}\).
\[\det e^T = \prod_i [e^T]_{ii} = \prod_i e^{T_{ii}} = e^{\sum_i T_{ii}} = e^{\mathrm{tr}\ T}\]Since \(A\) and \(T\) has the same set of eigenvalues, we have \(\mathrm{tr}\ A=\mathrm{tr}\ T\) Therefore
\[\det e^A = \det e^T = e^{\mathrm{tr}\ T} = e^{\mathrm{tr}\ A}\]
By the claim we just proved, one can show that \(\det W(t) = e^{\int_0^t \mathrm{tr}\ A(s)\ ds}\det W(0)\).
\[\det W(t) = \det W(t) = e^{\mathrm{tr}\int_0^t A(s)\ ds} \cdot \det W(0) = e^{\int_0^t \mathrm{tr}\ A(s)\ ds} \cdot \det W(0)\]
