midterm1.tex

%!TEX root = ./main.tex
\thispagestyle{empty}
\setlist[enumerate,2]{leftmargin = 2cm}
\section*{Additional Materials}
\subsection*{Midterm 1}
\begin{enumerate}
    \item $\li vm$ be linear independent in $V$, find $\dim \spa (v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n, v_n - v_1)$. 
    \begin{proof}
        Let $U = \spa ( v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n, v_n - v_1 )$
        Notice that $v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n, v_n - v_1$ is linearly independent, we guess that $\lb v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n, v_n - v_1 \rb$ is a basis for $U$. 
        \begin{enumerate} [label = \textit{Step \arabic*.}]
                \item We want to show that $ v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n$ is linearly independent. Suppose the equation  $a_1(v_1 - v_2) + a_2(v_2 - v_3) + \cdots + a_{n-1} (v_{n-1} - v_n) = 0\\ \implies a_1 v_1  + (a_2 - a_1) v_2 + \cdots + (a_{n-1} - a_n)v_{n-1} - a_{n-1} v_n = 0$. Since $\li vn$ is linearly independent, we know that $a_1 = a_2 - a_1 = \cdots = a_{n - 2} - a_{n_2} = a_{n-1} = 0 \implies \\ a_1 = a_2 = \cdots = a_{n-1} = 0$. Hence linear independence.
                \item We want to show that $U = \spa \lb  v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n \rb$ 
                \begin{align*}
                    U &= c_1(v_1 - v_2) + c_2(v_2 - v_3) + \cdots + c_n(v_n - v_1) \\
                    &= c_1(v_1 - v_2) + \cdots + c_n(v_{n-1} - v_n) + c_n[-(v_1 - v_2) - \cdots - (v_{n-1} - v_n)] \\
                    &= (c_1 - c_n)(v_1 - v_2) + \cdots + (c_{n - 1} - c_n)(v_{n-1}v_n) \\
                    &= \spa \lb  v_1 - v_2, v_2 - v_3, \ldots, v_{n-1} - v_n \rb
                \end{align*}
        \end{enumerate}
    \end{proof}
    \begin{proof}[Alternative Proof.]
        Consider $T \in \b F^n \to \b F^n$ be defined as \[T(v_1) = v_1 - v_2, T(v_2) = v_2 - v_3, \ldots, T(v_n) = v_n - v_1\] We can see that $\text{range } T = \spa (v_1 - v_2, v_2 - v_3, \ldots, v_n - v_1)$. Write everything in terms of matrix notation: 
        \[\bml 
        1 & 0 & \cdots & 0 & -1 \\
        -1 & 1 & \cdots & 0 & 0 \\
        \vdots & \vdots & \ddots & \vdots & \vdots  \\
        0 & 0 & \cdots & 1 & 0 \\
        0 & 0 & \cdots & -1 & 1 \\
        \bmr\] row reduce the matrix and we can see that $\dim \text{range } T = n - 1$.
    \end{proof}
    \item Let $W_1  = \lb (x_1, x_2, x_3, x_4 ) \in \b R^4 : x_1 + x_2 + x_3 + x_4 = 0 \rb, W_2 = \lb (x_1, x_2, x_3, x_4 ) \in \b R^4 : x_3  + x_4 \rb$. 
    \setlist[enumerate,2]{leftmargin = 1cm}
    \begin{enumerate}[label = \roman*.]
        \item Show that $W_1, W_2$ are subspace of $\b R^4$.
        \item Is $W_1 + W_2$ a direct sum?
        \item Compute $\dim (W_1 + W_2)$.
    \end{enumerate}
    \item Let $V = \lb f: \b R \to \b R \ | \ f^{(i)} \text{ exists for every } i \rb$, define $T_j$ as $T_j = f(x) \mapsto x^{j} f^{(j)}(x)$ for $j$ in $1, 2, \ldots, n$. Prove or Disprove: $T_1, T_2, \ldots, T_n$ is linearly independent.
    \begin{proof}
        We claim that $\li Tn$ is linearly independent. \\ 
        Let $\alpha_0 T_0 + \alpha_1 T_1 + \cdots + \alpha_n T_n$ be the zero map. We want to show that this is only possible if and only if $\alpha_0 = \alpha_1 = \cdots = \alpha_n = 0$.
        Let $f(x) = 1$. We can see that $T_0 f(x) = 1$ and $T_1 f(x) x, = T_2 f(x) = \cdots = T_n f(x) = 0$, therefore $\alpha_0 = 0$. Using a similar logic on $f(x) = x, x^2, \ldots, x^n$, we can conclude that $\alpha_0 = \alpha_1 = \cdots = \alpha_n = 0$. Hence $T_0, T_1, \ldots T_n$ are linearly independent.
    \end{proof}
    \item  
\end{enumerate}