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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.
- \(A = \left[ \begin{array}{rr} 2 & -4 \\ -3 & 3 \end{array}\right]\)
- \(A = \left[ \begin{array}{rr} 5 & 2 \\ -3 & -2 \end{array}\right]\)
- \(A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right]\)
- \(A = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 0 \end{array}\right]\)
- Answer
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- 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.
- \(A = \left[ \begin{array}{rr} 1 & 1 \\ 1 & 0 \end{array}\right]\)
- \(A = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 0 \end{array}\right]\)
- \(A_{1} = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 0 \end{array}\right]\), \(Q_{1} = \frac{1}{\sqrt{10}}\left[ \begin{array}{rr} 3 & -1 \\ 1 & 3 \end{array}\right]\), \(R_{1} = \frac{1}{\sqrt{10}}\left[ \begin{array}{rr} 10 & 3 \\ 0 & -1 \end{array}\right]\)
- Answer
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- \(A_{2} = \frac{1}{10}\left[ \begin{array}{rr} 33 & -1 \\ -1 & -3 \end{array}\right]\),
\(Q_{2} = \frac{1}{\sqrt{1090}}\left[ \begin{array}{rr} 33 & 1 \\ -1 & 33 \end{array}\right]\),
\(R_{2} = \frac{1}{\sqrt{1090}}\left[ \begin{array}{rr} 109 & -3 \\ 0 & -10 \end{array}\right]\)
\(A_{3} = \frac{1}{109}\left[ \begin{array}{rr} 360 & 1 \\ 1 & -33 \end{array}\right]\)
\({} = \left[ \begin{array}{rr} 3.302775 & 0.009174 \\ 0.009174 & -0.302775 \end{array}\right]\)
- \(A_{2} = \frac{1}{10}\left[ \begin{array}{rr} 33 & -1 \\ -1 & -3 \end{array}\right]\),
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.
- Answer
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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 5.5.5), 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 \(A_{k}\).
This page titled 8.5E: Computing Eigenvalues Exercises was last modified on Sat, 17 Jan 2026 00:22:41 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by W. Keith Nicholson.
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