example.tex

\documentclass[12pt]{homework}

\title{Matrix Analysis \quad Homework 6}
\author{Tao}
\email{tao@example.com}
\institute{School of Mathematics}

\begin{document}

\maketitle

\section*{1.2.P15} 
Suppose that $\mathbf{A}(t) \in M_n$ is a given continuous matrix-valued function and each of the vector valued functions $x_1(t) , \cdots, x_n(t) \in \mathbb{C}^n$ satisfies the system of ordinary differential equations
\begin{equation*}
    x^\prime _j (t) = \mathbf{A}(t) x _j (t).
\end{equation*}

Let $\mathbf{X}(t) = [x_1(t) \; \cdots \; x_n(t)]$ and let $W(t) = \det \mathbf{X}(t)$. Use (0.8.10) and (0.8.2.11) and provide details for the following argument:
\begin{equation*}
    \begin{aligned}
        W^{\prime}(t) &=\sum_{j=1}^{n} \operatorname{det}\left(\mathbf{X}(t) \xleftarrow[j]{} x_{j}^{\prime}(t)\right)=\operatorname{tr}\left[\operatorname{det}\left(\mathbf{X}(t) \xleftarrow[i]{} x_{j}^{\prime}(t)\right)\right]_{i, j=1}^{n} \\
        &=\operatorname{tr}\left((\operatorname{adj} \mathbf{X}(t)) \mathbf{X}^{\prime}(t)\right)=\operatorname{tr}((\operatorname{adj} \mathbf{X}(t)) \mathbf{A}(t) \mathbf{X}(t))=W(t) \operatorname{tr} \mathbf{A}(t).
    \end{aligned}
\end{equation*}
Thus, $W(t)$ satisfies the scalar differential equation $W^\prime (t) = \operatorname{tr} \mathbf{A}(t)W(t)$, whose solution is \textit{Abel’s formula} for the \textit{Wronskian}
\begin{equation*}
    W(t)=W\left(t_{0}\right) e^{\int_{t_{0}}^{t} \operatorname{tr} \mathbf{A}(s) \mathrm{d} s}.
\end{equation*}
Conclude that if the vectors $x_1(t) , \cdots, x_n(t)$ are linearly independent for $t=t_0$, then they are linearly independent for all $t$. How did you use the identity $\operatorname{tr} (\mathbf{B}\mathbf{C}) = \operatorname{tr} (\mathbf{C}\mathbf{B})$ (1.2.P2)?

\begin{solution}
    \begin{align*}
        W^\prime (t) &= \frac{d}{dt} \det \mathbf{X}(t) \\
        \overset{\text{(0.8.10)}}&{=} \sum_{j=1}^{n} \operatorname{det}\left(\mathbf{X}(t) \xleftarrow[j]{} x_{j}^{\prime}(t)\right) \\
        &= \operatorname{tr}\left[\operatorname{det}\left(\mathbf{X}(t) \xleftarrow[i]{} x_{j}^{\prime}(t)\right)\right]_{i, j=1}^{n} \\
        \overset{\text{(0.8.2.11)}}&{=} \operatorname{tr}\left((\operatorname{adj} \mathbf{X}(t)) \mathbf{X}^{\prime}(t)\right) \\
        \overset{x^\prime _j (t) = \mathbf{A}(t) x _j (t)}&{=} \operatorname{tr}((\operatorname{adj} \mathbf{X}(t)) \mathbf{A}(t) \mathbf{X}(t)) \\
        \overset{\operatorname{tr} (\mathbf{B}\mathbf{C})  = \operatorname{tr} (\mathbf{C}\mathbf{B})}&{=}  \operatorname{tr}( \mathbf{X}(t) (\operatorname{adj} \mathbf{X}(t)) \mathbf{A}(t))\\
        &= \det \mathbf{X}(t) \operatorname{tr}  \mathbf{A}(t) = W(t) \operatorname{tr} \mathbf{A}(t).
    \end{align*}

    If the vectors $x_1(t) , \cdots, x_n(t)$ are linearly independent for $t=t_0$, $W(t) = \det \mathbf{X}(t) \neq 0$.
    Hence $W(t)=W\left(t_{0}\right) e^{\int_{t_{0}}^{t} \operatorname{tr} \mathbf{A}(s) \mathrm{d} s} \neq 0$.
    Therefore $x_1(t) , \cdots, x_n(t)$ are linearly independent for all $t$.
\end{solution}

\end{document}