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5.11.1.6: Supplementary Exercises for Chapter 6

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    Exercise \(\PageIndex{1}\)

    (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 wronskian \(w(x)\) is nonzero for some \(x\), where

    \[ w(x) = \det \def\arraystretch{1.3} \left[ \begin{array}{ccc} f_1(x) & f_2(x) & f_3(x) \\ f_1^\prime(x) & f_2^\prime(x) & f_3^\prime(x) \\ f_1^\prime \prime(x) & f_2^\prime \prime(x) & f_3^\prime \prime(x) \\ \end{array} \right] \nonumber \]

    Answer

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    Exercise \(\PageIndex{2}\)

    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.

    1. If \(A\) is invertible, show that \(\{A\mathbf{v}_{1}, A\mathbf{v}_{2}, \dots, A\mathbf{v}_{n}\}\) is a basis of \(\mathbb{R}^n\).
    2. 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.
    Answer
    1. 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.
    Exercise \(\PageIndex{3}\)

    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}

    Answer

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    Exercise \(\PageIndex{4}\)

    Show that \(\text{null} A = \text{null}(A^{T}A)\) for any real matrix \(A\).

    Answer

    We have \(\text{null} A \subseteq \text{null}(A^{T}A)\) 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 \(A\mathbf{x} = \mathbf{0}\).

    Exercise \(\PageIndex{5}\)

    Let \(A\) be an \(m \times n\) matrix of rank \(r\). Show that \(dim \;(\text{null} A) = n - r\) (Theorem~\ref{thm:015672}) as follows. Choose a basis \(\{\mathbf{x}_{1}, \dots, \mathbf{x}_{k}\}\) of \(\text{null} A\) 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 \(\text{col} A\).

    Answer

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    This page titled 5.11.1.6: Supplementary Exercises for Chapter 6 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by W. Keith Nicholson via source content that was edited to the style and standards of the LibreTexts platform.