Chapter 5: Q5.2-24E (page 267)
Question: Show that if \(A\) and \(B\) are similar, then \(\det A = \det B\).
Short Answer
It is proved that \({\rm{det}}A = {\rm{det}}B\).
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Chapter 5: Q5.2-24E (page 267)
Question: Show that if \(A\) and \(B\) are similar, then \(\det A = \det B\).
It is proved that \({\rm{det}}A = {\rm{det}}B\).
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Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)
Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).
Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).
14. \(\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\)
Compute the quantities in Exercises 1-8 using the vectors
\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)
2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)
Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).
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