8.5E: Computing Eigenvalues Exercises
- Page ID
- 132841
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercises for 1
solutions
2
In each case, find the exact eigenvalues and determine corresponding eigenvectors. Then start with \(\mathbf{x}_{0} = \left[ \begin{array}{rr} 1 \\ 1 \end{array}\right]\) and compute \(\mathbf{x}_{4}\) and \(r_{3}\) using the power method.
\(A = \left[ \begin{array}{rr} 2 & -4 \\ -3 & 3 \end{array}\right]\) \(A = \left[ \begin{array}{rr} 5 & 2 \\ -3 & -2 \end{array}\right]\) \(A = \left[ \begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right]\) \(A = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 0 \end{array}\right]\)
- Eigenvalues \(4\), \(-1\); eigenvectors \(\left[ \begin{array}{rr} 2 \\ -1 \end{array}\right]\), \(\left[ \begin{array}{rr} 1 \\ -3 \end{array}\right]\); \(\mathbf{x}_{4} = \left[ \begin{array}{rr} 409 \\ -203 \end{array}\right]\); \(r_{3} = 3.94\)
- Eigenvalues \(\lambda_{1} = \frac{1}{2}(3 + \sqrt{13})\), \(\lambda_{2} = \frac{1}{2}(3 - \sqrt{13})\); eigenvectors \(\left[ \begin{array}{c} \lambda_{1} \\ 1 \end{array}\right]\), \(\left[ \begin{array}{c} \lambda_{2} \\ 1 \end{array}\right]\); \(\mathbf{x}_{4} = \left[ \begin{array}{rr} 142 \\ 43 \end{array}\right]\); \(r_{3} = 3.3027750\) (The true value is \(\lambda_{1} = 3.3027756\), to seven decimal places.)
In each case, find the exact eigenvalues and then approximate them using the QR-algorithm.
\(A = \left[ \begin{array}{rr} 1 & 1 \\ 1 & 0 \end{array}\right]\) \(A = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 0 \end{array}\right]\)
- \(A_{1} = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 0 \end{array}\right]\), \(Q_{1} = \frac{1}{\sqrt{10}}\left[ \begin{array}{rr} 3 & -1 \\ 1 & 3 \end{array}\right]\), \(R_{1} = \frac{1}{\sqrt{10}}\left[ \begin{array}{rr} 10 & 3 \\ 0 & -1 \end{array}\right]\)
\(A_{2} = \frac{1}{10}\left[ \begin{array}{rr} 33 & -1 \\ -1 & -3 \end{array}\right]\),
\(Q_{2} = \frac{1}{\sqrt{1090}}\left[ \begin{array}{rr} 33 & 1 \\ -1 & 33 \end{array}\right]\),
\(R_{2} = \frac{1}{\sqrt{1090}}\left[ \begin{array}{rr} 109 & -3 \\ 0 & -10 \end{array}\right]\)
\(A_{3} = \frac{1}{109}\left[ \begin{array}{rr} 360 & 1 \\ 1 & -33 \end{array}\right]\)
\({} = \left[ \begin{array}{rr} 3.302775 & 0.009174 \\ 0.009174 & -0.302775 \end{array}\right]\)
Apply the power method to
\(A = \left[ \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]\), starting at \(\mathbf{x}_{0} = \left[ \begin{array}{rr} 1 \\ 1 \end{array}\right]\). Does it converge? Explain.
If \(A\) is symmetric, show that each matrix \(A_{k}\) in the QR-algorithm is also symmetric. Deduce that they converge to a diagonal matrix.
Use induction on \(k\). If \(k = 1\), \(A_{1} = A\). In general \(A_{k+1} = Q_{k}^{-1}A_{k}Q_{k} = Q_{k}^{T}A_{k}Q_{k}\), so the fact that \(A_{k}^{T} = A_{k}\) implies \(A_{k+1}^{T} = A_{k+1}\). The eigenvalues of \(A\) are all real (Theorem [thm:016145]), so the \(A_{k}\) converge to an upper triangular matrix \(T\). But \(T\) must also be symmetric (it is the limit of symmetric matrices), so it is diagonal.
Apply the QR-algorithm to
\(A = \left[ \begin{array}{rr} 2 & -3 \\ 1 & -2 \end{array}\right]\). Explain.
Given a matrix \(A\), let \(A_{k}\), \(Q_{k}\), and \(R_{k}\), \(k \geq 1\), be the matrices constructed in the QR-algorithm. Show that \(A_{k} = (Q_{1}Q_{2} \cdots Q_{k})(R_{k} \cdots R_{2}R_{1})\) for each \(k \geq 1\) and hence that this is a QR-factorization of \(A_{k}\).