4.2.1: Exercises 4.2
- Page ID
- 70775
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In Exercises \(\PageIndex{1}\) – \(\PageIndex{6}\), a matrix \(A\) is given. For each,
- Find the eigenvalues of \(A\), and for each eigenvalue, find an eigenvector.
- Do the same for \(A^{T}\).
- Do the same for \(A^{−1}\)
- Find \(\text{tr}(A)\).
- Find \(\text{det}(A)\). Use Theorem 4.2.1 to verify your results.
\(\left[\begin{array}{cc}{0}&{4}\\{-1}&{5}\end{array}\right]\)
- Answer
-
- \(\lambda_{1}=1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{4}\\{1}\end{array}\right];\)
\(\lambda_{2}=4\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{1}\\{1}\end{array}\right]\) - \(\lambda_{1}=1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{1}\end{array}\right];\)
\(\lambda_{2}=4\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-1}\\{4}\end{array}\right]\) - \(\lambda_{1}=1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{1}\\{1}\end{array}\right];\)
\(\lambda_{2}=1\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{4}\\{1}\end{array}\right]\) - \(5\)
- \(4\)
- \(\lambda_{1}=1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{4}\\{1}\end{array}\right];\)
\(\left[\begin{array}{cc}{-2}&{-14}\\{-1}&{3}\end{array}\right]\)
- Answer
-
- \(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{7}\\{1}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-2}\\{1}\end{array}\right]\) - \(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{1}\\{2}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-1}\\{7}\end{array}\right]\) - \(\lambda_{1}=-1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{7}\\{1}\end{array}\right];\)
\(\lambda_{2}=1/5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-2}\\{1}\end{array}\right]\) - \(1\)
- \(-20\)
- \(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{7}\\{1}\end{array}\right];\)
\(\left[\begin{array}{cc}{5}&{30}\\{-1}&{-6}\end{array}\right]\)
- Answer
-
- \(\lambda_{1}=-1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\end{array}\right];\)
\(\lambda_{2}=0\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-6}\\{1}\end{array}\right]\) - \(\lambda_{1}=-1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{1}\\{6}\end{array}\right];\)
\(\lambda_{2}=0\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{1}\\{5}\end{array}\right]\) - \(A\) is not invertible.
- \(-1\)
- \(0\)
- \(\lambda_{1}=-1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\end{array}\right];\)
\(\left[\begin{array}{cc}{-4}&{72}\\{-1}&{13}\end{array}\right]\)
- Answer
-
- \(\lambda_{1}=4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{9}\\{1}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{8}\\{1}\end{array}\right]\) - \(\lambda_{1}=4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{8}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-1}\\{9}\end{array}\right]\) - \(\lambda_{1}=1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{9}\\{1}\end{array}\right];\)
\(\lambda_{2}=1/5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{8}\\{1}\end{array}\right]\) - \(9\)
- \(20\)
- \(\lambda_{1}=4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{9}\\{1}\end{array}\right];\)
\(\left[\begin{array}{ccc}{5}&{-9}&{0}\\{1}&{-5}&{0}\\{2}&{4}&{3}\end{array}\right]\)
- Answer
-
- \(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-7}\\{-7}\\{6}\end{array}\right];\)
\(\lambda_{2}=3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=4\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{9}\\{1}\\{22}\end{array}\right]\) - \(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{9}\\{0}\end{array}\right];\)
\(\lambda_{2}=3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-20}\\{26}\\{7}\end{array}\right]\)
\(\lambda_{3}=4\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{-1}\\{1}\\{0}\end{array}\right]\) - \(\lambda_{1}=-1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-7}\\{-7}\\{6}\end{array}\right];\)
\(\lambda_{2}=1/3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=1/4\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{9}\\{1}\\{22}\end{array}\right]\) - \(3\)
- \(-48\)
- \(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-7}\\{-7}\\{6}\end{array}\right];\)
\(\left[\begin{array}{ccc}{0}&{25}&{0}\\{1}&{0}&{0}\\{1}&{1}&{-3}\end{array}\right]\)
- Answer
-
- \(\lambda_{1}=-5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\\{2}\end{array}\right];\)
\(\lambda_{2}=-3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=5\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{20}\\{4}\\{3}\end{array}\right]\) - \(\lambda_{1}=-5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{5}\\{0}\end{array}\right];\)
\(\lambda_{2}=-3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{1}\\{-11}\\{8}\end{array}\right]\)
\(\lambda_{3}=5\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{1}\\{5}\\{0}\end{array}\right]\) - \(\lambda_{1}=-1/5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\\{2}\end{array}\right];\)
\(\lambda_{2}=-1/3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=1/5\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{20}\\{4}\\{3}\end{array}\right]\) - \(-3\)
- \(75\)
- \(\lambda_{1}=-5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\\{2}\end{array}\right];\)