3.3E: Eigenvalues and Eigenvectors Exercises
- Page ID
- 132811
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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 each case find the characteristic polynomial, eigenvalues, eigenvectors, and (if possible) an invertible matrix \(P\) such that \(P^{-1}AP\) is diagonal.
- \(A = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 2 \end{array}\right]\)
- \(A = \left[ \begin{array}{rr} 2 & -4 \\ -1 & -1 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 7 & 0 & -4 \\ 0 & 5 & 0 \\ 5 & 0 & -2 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 1 & 1 & -3 \\ 2 & 0 & 6 \\ 1 & -1 & 5 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 1 & -2 & 3 \\ 2 & 6 &-6 \\ 1 & 2 & -1 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 0 & 1 & 0 \\ 3 & 0 & 1 \\ 2 & 0 & 0 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 3 & 1 & 1 \\ -4 & -2 & -5 \\ 2 & 2 & 5 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & -1 & 2 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu \end{array}\right]\), \(\lambda \neq \mu\)
- Answer
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- \((x-3)(x+2); 3; -2\); \(\left[ \begin{array}{r} 4 \\ -1 \end{array} \right], \left[ \begin{array}{r} 1 \\ 1 \end{array} \right]\); \(P = \left[ \begin{array}{rr} 4 & 1 \\ -1 & 1 \end{array} \right]\); \(P^{-1}AP = \left[ \begin{array}{rr} 3 & 0 \\ 0 & -2 \end{array}\right].\)
- \((x-2)^3 ; 2 ; \left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right], \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array}\right]\); No such \(P\); Not diagonalizable.
- \((x+1)^2(x-2) ; -1, -2; \left[ \begin{array}{r} -1 \\ 1 \\ 2 \end{array}\right], \left[ \begin{array}{r} 1 \\ 2 \\ 1 \end{array}\right]\); No such \(P\); Not diagonalizable. Note that this matrix and the matrix in Example 3.4.2 have the same characteristic polynomial, but that matrix is diagonalizable.
- \((x-1)^2(x-3) ; 1, 3; \left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right], \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array}\right]\) No such \(P\); Not diagonalizable.
Consider a linear dynamical system \(\mathbf{v}_{k+1} = A\mathbf{v}_{k}\) for \(k \geq 0\). In each case approximate \(\mathbf{v}_{k}\) using Theorem 3.5.1.
- \(A = \left[ \begin{array}{rr} 2 & 1 \\ 4 & -1 \end{array}\right], \mathbf{v}_0 = \left[ \begin{array}{r} 1 \\ 2 \end{array}\right]\)
- \(A = \left[ \begin{array}{rr} 3 & -2 \\ 2 & -2 \end{array}\right], \mathbf{v}_0 = \left[ \begin{array}{r} 3 \\ -1 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 2 & 3 \\ 1 & 4 & 1 \end{array}\right], \mathbf{v}_0 = \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 1 & 3 & 2 \\ -1 & 2 & 1 \\ 4 & -1 & -1 \end{array}\right], \mathbf{v}_0 = \left[ \begin{array}{r} 2 \\ 0 \\ 1 \end{array}\right]\)
- Answer
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- \(V_k = \frac{7}{3} 2^k \left[ \begin{array}{r} 2 \\ 1 \end{array}\right]\)
- \(V_k = \frac{3}{2} 3^k \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array}\right]\)
Show that \(A\) has \(\lambda = 0\) as an eigenvalue if and only if \(A\) is not invertible.
Let \(A\) denote an \(n \times n\) matrix and put \(A_{1} = A - \alpha I\), \(\alpha\) in \(\mathbb{R}\). Show that \(\lambda\) is an eigenvalue of \(A\) if and only if \(\lambda -\alpha\) is an eigenvalue of \(A_{1}\). (Hence, the eigenvalues of \(A_{1}\) are just those of \(A\) “shifted” by \(\alpha\).) How do the eigenvectors compare?
- Answer
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\(A\mathbf{x} = \lambda\mathbf{x}\) if and only if \((A - \alpha I)\mathbf{x} = (\lambda - \alpha)\mathbf{x}\). Same eigenvectors.
Show that the eigenvalues of \(\left[ \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right]\) are \(e^{i\theta}\) and \(e^{-i\theta}\). (See Appendix A)
Find the characteristic polynomial of the \(n \times n\) identity matrix \(I\). Show that \(I\) has exactly one eigenvalue and find the eigenvectors.
Given \(A = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]\) show that:
- \(c_{A}(x) = x^{2} - tr \; Ax + \det A\), where \(tr \; A = a + d\) is called the trace of \(A\).
- The eigenvalues are \(\frac{1}{2} \left[ (a+d) \pm \sqrt{(a-b)^2 + 4bc}\right]\).
In each case, find \(P^{-1}AP\) and then compute \(A^{n}\).
- \(A = \left[ \begin{array}{rr} 6 & -5 \\ 2 & -1 \end{array}\right], P = \left[ \begin{array}{rr} 1 & 5 \\ 1 & 2 \end{array}\right]\)
- \(A = \left[ \begin{array}{rr} -7 & -12 \\ 6 & -10 \end{array}\right], P = \left[ \begin{array}{rr} -3 & 4 \\ 2 & -3 \end{array}\right]\)
[Hint: \((PDP^{-1})^{n} = PD^{n}P^{-1}\) for each \(n = 1, 2, \dots\).]
- Answer
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- \(P^{-1}AP = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array}\right]\), so \(A^n = P \left[ \begin{array}{rr} 1 & 0 \\ 0 & 2^n \end{array}\right] P^{-1} = \left[ \begin{array}{cc} 9 - 8 \cdot 2^n & 12(1-2^n) \\ 6(2^n-1) & 9\cdot 2^n - 8 \end{array}\right]\)