5.11.1.6: Supplementary Exercises for Chapter 6

(Requires calculus) Let $V$ denote the space of all functions $f : \mathbb{R} \to \mathbb{R}$ for which the derivatives $f^\prime$ and $f^\prime \prime$ exist. Show that $f_{1}$, $f_{2}$, and $f_{3}$ in $V$ are linearly independent provided that their \textbf{wronskian}\index{wronskian} $w(x)$ is nonzero for some $x$, where \begin{equation*} w(x) = \det \def\arraystretch{1.3} \left[ \begin{array}{ccc} f_1(x) & f_2(x) & f_3(x) \\ f_1^\prime(x) & f_2^\prime(x) & f_3^\prime(x) \\ f_1^\prime \prime(x) & f_2^\prime \prime(x) & f_3^\prime \prime(x) \\ \end{array} \right] \end{equation*} \end{supex} \begin{supex} Let $\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}$ be a basis of $\mathbb{R}^n$ (written as columns), and let $A$ be an $n \times n$ matrix.

\item If $A$ is invertible, show that $\{A\mathbf{v}_{1}, A\mathbf{v}_{2}, \dots, A\mathbf{v}_{n}\}$ is a basis of $\mathbb{R}^n$.

\item If $\{A\mathbf{v}_{1}, A\mathbf{v}_{2}, \dots, A\mathbf{v}_{n}\}$ is a basis of $\mathbb{R}^n$, show that $A$ is invertible.

{1} \item If $YA = 0$, $Y$ a row, we show that $Y = 0$; thus $A^{T}$ (and hence $A$) is invertible. Given a column $\mathbf{c}$ in $\mathbb{R}^n$ write $\mathbf{c} = \displaystyle \sum_{i}r_i(A\mathbf{v}_i)$ where each $r_{i}$ is in $\mathbb{R}$. Then $Y \mathbf{c} = \displaystyle \sum_{i}r_iYA\mathbf{v}_i$, so $Y = YI_n = Y \left[ \begin{array}{cccc} \mathbf{e}_1 & \mathbf{e}_2 & \cdots & \mathbf{e}_n \end{array} \right] = \left[ \begin{array}{cccc} Y\mathbf{e}_1 & Y\mathbf{e}_2 & \cdots & Y\mathbf{e}_n \end{array} \right] = \left[ \begin{array}{cccc} 0 & 0 & \cdots & 0 \end{array} \right] = 0$, as required. \end{enumerate} \end{supsol} \end{supex} \columnbreak \begin{supex} If $A$ is an $m \times n$ matrix, show that $A$ has rank $m$ if and only if col $A$ contains every column of $I_{m}$. \end{supex} \begin{supex} Show that $\func{null} A = \func{null}(A^{T}A)$ for any real matrix $A$. \begin{supsol} We have $\func{null} A \subseteq \func{null}(A^{T}A)$ because $A\mathbf{x} = \mathbf{0}$ implies $(A^{T}A)\mathbf{x} = \mathbf{0}$. Conversely, if $(A^{T}A)\mathbf{x} = \mathbf{0}$, then $\| A\mathbf{x}\|^{2} = (A\mathbf{x})^{T}(A\mathbf{x}) = \mathbf{x}^{T}A^{T}A\mathbf{x} = 0$. Thus $A\mathbf{x} = \mathbf{0}$. \end{supsol} \end{supex} \begin{supex} Let $A$ be an $m \times n$ matrix of rank $r$. Show that $dim \;(\func{null} A) = n - r$ (Theorem~\ref{thm:015672}) as follows. Choose a basis $\{\mathbf{x}_{1}, \dots, \mathbf{x}_{k}\}$ of $\func{null} A$ and extend it to a basis $\{\mathbf{x}_{1}, \dots, \mathbf{x}_{k}, \mathbf{z}_{1}, \dots, \mathbf{z}_{m}\}$ of $\mathbb{R}^n$. Show that $\{A\mathbf{z}_{1}, \dots, A\mathbf{z}_{m}\}$ is a basis of $\func{col} A$. \end{supex} \end{multicols}