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Chapter 12: Problem 2

Assume that \(\mathbf{X}\) is an \(n \times p\) matrix. Then the kernel of \(\mathbf{X}\) is defined to be the space \(\operatorname{ker}(\mathbf{X})=\\{\mathbf{b}: \mathbf{X} \mathbf{b}=\mathbf{0}\\}\) (a) Show that \(\operatorname{ker}(\mathbf{X})\) is a subspace of \(R^{p}\). (b) The dimension of \(\operatorname{ker}(\mathbf{X})\) is called the nullity of \(\mathbf{X}\) and is denoted by \(\nu(\mathbf{X})\). Let \(\rho(\mathbf{X})\) denote the rank of \(\mathbf{X}\). A fundamental theorem of linear algebra says that \(\rho(\mathbf{X})+\nu(\mathbf{X})=p .\) Use this to show that if \(\mathbf{X}\) has full column rank, then \(\operatorname{ker}(\mathbf{X})=\\{\mathbf{0}\\}\)

Short Answer

Expert verified
The kernel, \(\operatorname{ker}(\mathbf{X})\), is a subspace of \(R^p\). When a matrix \(\mathbf{X}\) has full column rank, the only solution to \(\mathbf{X}\mathbf{b}=\mathbf{0}\), which implies that \(\operatorname{ker}(\mathbf{X})= \{\mathbf{0}\}\).

Step by step solution

01

Prove that kernel is a subspace

A subset W of a vector space V is a subspace if it satisfies three properties:\n(a) The zero vector of V is in W.\n(b) W is closed under vector addition.\n(c) W is closed under scalar multiplication.\n\nLet's denote that \(\mathbf{z}\) is the zero vector of \(R^p\), and \(\mathbf{b_1}\), \(\mathbf{b_2}\) are vectors from \(\operatorname{ker}(\mathbf{X})\), and \(r\) is any scalar.\n\n(a) \(\mathbf{X}\mathbf{z} = \mathbf{0}\)\n(b) \(\mathbf{X}(\mathbf{b_1} + \mathbf{b_2}) = \mathbf{X}\mathbf{b_1} + \mathbf{X}\mathbf{b_2} = \mathbf{0} + \mathbf{0} = \mathbf{0}\)\n(c) \(\mathbf{X}(r\mathbf{b_1}) = r(\mathbf{X}\mathbf{b_1}) = r\mathbf{0} = \mathbf{0}\)\n\nAs these three properties are satisfied, \(\operatorname{ker}(\mathbf{X})\) is a subspace of \(R^p\).
02

Prove that if \(\mathbf{X}\) has full column rank, then \(\operatorname{ker}(\mathbf{X})= \{\mathbf{0}\}\)

Matrix \(\mathbf{X}\) having full column rank means that \(\rho(\mathbf{X})=p\). If we substitute this into the rank-nullity theorem \(\rho(\mathbf{X})+\nu(\mathbf{X})=p\), we have \(p + \nu(\mathbf{X}) = p\), which implies that \(\nu(\mathbf{X}) = 0\). Hence, the dimension of the kernel, also called the nullity, is 0. Therefore, it contains only the zero vector. As a result, \(\operatorname{ker}(\mathbf{X})= \{\mathbf{0}\}\).

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Most popular questions from this chapter

Often influence functions are derived by differentiating implicitly the defining equation for the functional at the contaminated cdf \(F_{y, \epsilon}(t),(12.1 .18)\). Consider the mean functional with the defining equation (12.1.15). Using the linearity of the differential, first show that the defining equation at the \(\operatorname{cdf} F_{y, \epsilon}(t)\) can be expressed as $$ \begin{aligned} 0=\int_{-\infty}^{\infty}\left[t-T\left(F_{y, \epsilon}\right)\right] d F_{y, \epsilon}(t)=&(1-\epsilon) \int_{-\infty}^{\infty}\left[t-T\left(F_{y, \epsilon}\right)\right] f_{Y}(t) d t \\ &+\epsilon \int_{-\infty}^{\infty}\left[t-T\left(F_{y, \epsilon}\right)\right] d_{\Delta}(t) \end{aligned} $$ Recall that we want \(\partial T\left(F_{y, \epsilon}\right) / \partial \epsilon\). Obtain this by differentiating implicitly the above equation with respect to \(\epsilon\).

Establish the identity $$ \|\mathbf{v}\|_{W}=\frac{\sqrt{3}}{2(n+1)} \sum_{i=1}^{n} \sum_{j=1}^{n}\left|v_{i}-v_{j}\right| $$ for all \(\mathbf{v} \in R^{n}\). Thus we have shown that $$ \widehat{\beta}_{W}=\operatorname{Argmin} \sum_{i=1}^{n} \sum_{j=1}^{n}\left|\left(y_{i}-y_{j}\right)-\beta\left(x_{c i}-x_{c j}\right)\right| . $$ Note that the formulation of \(\widehat{\beta}_{W}\) given in expression \((12.2 .29)\) allows an easy way to compute the Wilcoxon estimate of slope by using an \(L_{1}\), (least absolute deviations), routine. This was used in the cited article by Terpstra, et al. for their \(\mathrm{R}\) or S-PLUS functions which compute the Wilcoxon fit.

This exercise assumes that the package S-PLUS is available or a package which computes an \(L_{1}\) fit. Consider the simple data set $$ \begin{array}{|l|r|r|r|r|r|r|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Y & 18.1 & 6.5 & 10.1 & 14.9 & 21.9 & 22.9 \\ \hline \end{array} $$

To complete the proof of Theorem 12.3.3, we need to show that the LS estimator \(\widehat{\eta}=\mathbf{H} \mathbf{Y}\) is unique. Suppose that \(\widehat{\boldsymbol{\eta}}_{2} \in V\) is also a LS solution. (a) Show that $$ \left\|\hat{\eta}-\widehat{\eta}_{2}\right\|^{2}=2\|(\mathbf{I}-\mathbf{H}) \mathbf{Y}\|^{2}-2 \mathbf{Y}^{\prime}(\mathbf{I}-\mathbf{H})\left(\mathbf{Y}-\widehat{\eta}_{2}\right) $$ (b) Now show that the right side of this last expression is 0 and, hence, establish the uniqueness.

Consider the linear model (12.3.2) with the design matrix given by \(\mathbf{X}=\) \(\left[1 \mathrm{c}_{1} \cdots \mathbf{c}_{p}\right] .\) Assume that the columns of the design matrix \(\mathbf{X}\) are orthogonal. Show that the LS estimate of \(\mathbf{b}\) is given by \(\widehat{\mathbf{b}}^{\prime}=\left(\bar{Y}, \widehat{b}_{1}, \ldots, \widehat{b}_{p}\right)\) where \(\widehat{b}_{j}\) is the LS estimate of the simple regression model \(Y_{i}=b_{j} c_{i j}+\varepsilon_{i}\), for \(j=1, \ldots, p .\) That is, the LS multiple regression estimator in this case is found by the individual simple LS regression estimators.
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