Chapter 8: Problem 55
\(\operatorname{det} \mathbf{A}=0,\) so the system has a nontrivial solution.
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Chapter 8: Problem 55
\(\operatorname{det} \mathbf{A}=0,\) so the system has a nontrivial solution.
These are the key concepts you need to understand to accurately answer the question.
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If we take \(\mathbf{K}_{1}=\left(\begin{array}{l}0 \\ 1 \\\ 1\end{array}\right)\) as in Example 4 in the text then we look for a vector \(\mathbf{K}_{2}=\left(\begin{array}{l}a \\ b \\ c\end{array}\right)\) such that \(1(a)+\) \(\frac{1}{4} b-\frac{1}{4} c=0\) and \(\mathbf{K}_{1} \cdot \mathbf{K}_{2}=0\) or \(b+c=0 .\) The last equation implies \(c=-b\) so \(a+\frac{1}{4} b-\frac{1}{4}(-b)=a+\frac{1}{2} b=0\) If we let \(b=-2,\) then \(a=1\) and \(c=2,\) so a second eigenvector with eigenvalue -9 and orthogonal to \(\mathbf{K}_{1}\) is \(\mathbf{K}_{2}=\left(\begin{array}{r}1 \\ -2 \\ 2\end{array}\right)\)
We solve \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\left|\begin{array}{cc}4-\lambda & 8 \\ 0 & -5-\lambda\end{array}\right|=(\lambda-4)(\lambda+5)=0\). For \(\lambda_{1}=4\) we have \(\left(\begin{array}{rr|r}0 & 8 & 0 \\ 0 & -9 & 0\end{array}\right) \Longrightarrow\left(\begin{array}{ll|l}0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right)\) so that \(k_{2}=0 .\) If \(k_{1}=1\) then \(\mathbf{K}_{1}=\left(\begin{array}{l}1 \\ 0\end{array}\right) .\) For \(\lambda_{2}=-5\) we have \(\left(\begin{array}{cc|c}9 & 8 & 0 \\ 0 & 0 & 0\end{array}\right) \Longrightarrow\left(\begin{array}{cc|c}1 & \frac{8}{9} & 0 \\ 0 & 0 & 0\end{array}\right)\) so that \(k_{1}=-\frac{8}{9} k_{2} .\) If \(k_{2}=9\) then \(\mathbf{K}_{2}=\left(\begin{array}{r}-8 \\\ 9\end{array}\right)\).
Distinct eigenvalues \(\lambda_{1}=-7, \lambda_{2}=4\) imply \(\mathbf{A}\) is diagonalizable. $$\mathbf{P}=\left(\begin{array}{cc} 13 & 1 \\ 2 & 1 \end{array}\right), \quad \mathbf{D}=\left(\begin{array}{rr} -7 & 0 \\ 0 & 4 \end{array}\right)$$
$$\mathbf{A}=\left(\mathbf{A}^{-1}\right)^{-1}=\left(\begin{array}{rr} -2 & 3 \\ 3 & -4 \end{array}\right)$$
$$\text { We have } \quad \mathbf{Y}^{T}=\left(\begin{array}{ccccccccc} 2 & 2.5 & 1 & 1.5 & 2 & 3.2 & 5 \end{array}\right) \text { and } \quad \mathbf{A}^{T}=\left(\begin{array}{ccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{array}\right).$$ $$\mathrm{Now} \quad \mathbf{A}^{T} \mathbf{A}=\left(\begin{array}{cc} 140 & 28 \\ 28 & 7 \end{array}\right) \quad \text { and } \quad\left(\mathbf{A}^{T} \mathbf{A}\right)^{-1}=\frac{1}{196}\left(\begin{array}{rr} 7 & -28 \\ -28 & 140 \end{array}\right)$$ $$\text { so } \mathbf{X}=\left(\mathbf{A}^{T} \mathbf{A}\right)^{-1} \mathbf{A}^{T} \mathbf{Y}=\left(\begin{array}{c} 0.407143 \\ 0.828571 \end{array}\right) \text { and the least squares line is } y=0.407143 x+0.828571.$$
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