3.4E: Diagonalization Exercises
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- 214767
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- If \(A = \left[ \begin{array}{rr} 1 & 3 \\ 0 & 2 \end{array} \right]\) and \(B = \left[ \begin{array}{rr} 2 & 0 \\ 0 & 1 \end{array}\right]\) verify that \(A\) and \(B\) are diagonalizable, but \(AB\) is not.
- If \(D = \left[ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right]\) find a diagonalizable matrix \(A\) such that \(D + A\) is not diagonalizable.
- Answer
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- \(A = \left[ \begin{array}{rr} 0 & 1 \\ 0 & 2 \end{array}\right]\)
If \(A\) is an \(n \times n\) matrix, show that \(A\) is diagonalizable if and only if \(A^{T}\) is diagonalizable.
If \(A\) is diagonalizable, show that each of the following is also diagonalizable.
- \(A^{n}\), \(n \geq 1\)
- \(kA\), \(k\) any scalar.
- \(p(A)\), \(p(x)\) any polynomial (Theorem 3.3.1)
- \(U^{-1}AU\) for any invertible matrix \(U\).
- \(kI + A\) for any scalar \(k\).
- Answer
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- \(PAP^{-1} = D\) is diagonal
- \(P^{-1}(kA)P = kD\) is diagonal
- \(Q(U^{-1}AU)Q = D\) where \(Q = PU\).
Give an example of two diagonalizable matrices \(A\) and \(B\) whose sum \(A + B\) is not diagonalizable.
- Answer
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\(\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]\) is not diagonalizable by Example 3.4.1. But \(\left[ \begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array}\right] = \left[ \begin{array}{rr} 2 & 1 \\ 0 & -1 \end{array}\right] + \left[ \begin{array}{rr} -1 & 0 \\ 0 & 2 \end{array}\right]\) where \(\left[ \begin{array}{rr} 2 & 1 \\ 0 & -1 \end{array}\right]\) has diagonalizing matrix \(P = \left[ \begin{array}{rr} 1 & -1 \\ 0 & 3 \end{array}\right]\) and \(\left[ \begin{array}{rr} -1 & 0 \\ 0 & 2 \end{array}\right]\) is already diagonal.
If \(A\) is diagonalizable and \(1\) and \(-1\) are the only eigenvalues, show that \(A^{-1} = A\).
If \(A\) is diagonalizable and \(0\) and \(1\) are the only eigenvalues, show that \(A^{2} = A\).
- Answer
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We have \(\lambda^{2} = \lambda\) for every eigenvalue \(\lambda\) (as \(\lambda = 0, 1\)) so \(D^{2} = D\), and so \(A^{2} = A\) as in Example 3.4.2.
If \(A\) is diagonalizable and \(\lambda \geq 0\) for each eigenvalue of \(A\), show that \(A = B^{2}\) for some matrix \(B\).
If \(P^{-1}AP\) and \(P^{-1}BP\) are both diagonal, show that \(AB = BA\). [Hint: Diagonal matrices commute.]
A square matrix \(A\) is called nilpotent if \(A^{n} = 0\) for some \(n \geq 1\). Find all nilpotent diagonalizable matrices. [Hint: Theorem 3.3.1.]
Let \(A\) be any \(n \times n\) matrix and \(r \neq 0\) a real number.
- Show that the eigenvalues of \(rA\) are precisely the numbers \(r\lambda\), where \(\lambda\) is an eigenvalue of \(A\).
- Show that \(c_{rA}(x) = r^n c_A\left( \frac{x}{r} \right)\).
- Answer
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- \(c_{rA} (x) =\det \left[ xI - rA \right]\) \({} = r^n \det \left[ \frac{x}{r}I-A \right] = r^n c_A \left[ \frac{x}{r} \right]\)
- If all rows of \(A\) have the same sum \(s\), show that \(s\) is an eigenvalue.
- If all columns of \(A\) have the same sum \(s\), show that \(s\) is an eigenvalue.
Let \(A\) be an invertible \(n \times n\) matrix.
- Show that the eigenvalues of \(A\) are nonzero.
- Show that the eigenvalues of \(A^{-1}\) are precisely the numbers \(1/\lambda\), where \(\lambda\) is an eigenvalue of \(A\).
- Show that \(c_{A^{-1}}(x) = \frac{(-x)^n}{\det A} c_A \left( \frac{1}{x} \right)\).
- Answer
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- If \(\lambda \neq 0\), \(A\mathbf{x} = \lambda\mathbf{x}\) if and only if \(A^{-1}\mathbf{x} = \frac{1}{\lambda}\mathbf{x}\). The result follows.
Suppose \(\lambda\) is an eigenvalue of a square matrix \(A\) with eigenvector \(\mathbf{x} \neq \mathbf{0}\).
- Show that \(\lambda^{2}\) is an eigenvalue of \(A^{2}\) (with the same \(\mathbf{x}\)).
- Show that \(\lambda^{3} - 2 \lambda + 3\) is an eigenvalue of \(A^{3} - 2A + 3I\).
- Show that \(p(\lambda)\) is an eigenvalue of \(p(A)\) for any nonzero polynomial \(p(x)\).
- Answer
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- \((A^{3} - 2A - 3I)\mathbf{x} = A^{3}\mathbf{x} - 2A\mathbf{x} + 3\mathbf{x} = \lambda^{3}\mathbf{x} - 2\lambda\mathbf{x} + 3\mathbf{x} = (\lambda^{3} - 2\lambda - 3)\mathbf{x}\).
If \(A\) is an \(n \times n\) matrix, show that \(c_{A^2}(x^{2}) = (-1)^{n}c_{A}(x)c_{A}(-x)\).
An \(n \times n\) matrix \(A\) is called nilpotent if \(A^{m} = 0\) for some \(m \geq 1\).
- Show that every triangular matrix with zeros on the main diagonal is nilpotent.
- If \(A\) is nilpotent, show that \(\lambda = 0\) is the only eigenvalue (even complex) of \(A\).
- Deduce that \(c_{A}(x) = x^{n}\), if \(A\) is \(n \times n\) and nilpotent.
- Answer
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- If \(A^{m} = 0\) and \(A\mathbf{x} = \lambda\mathbf{x}\), \(\mathbf{x} \neq \mathbf{0}\), then \(A^{2}\mathbf{x} = A(\lambda\mathbf{x}) = \lambda A\mathbf{x} = \lambda^{2}\mathbf{x}\). In general, \(A^{k}\mathbf{x} = \lambda^{k}\mathbf{x}\) for all \(k \geq 1\). Hence, \(\lambda^{m}\mathbf{x} = A^{m}\mathbf{x} = \mathbf{0}\mathbf{x} = \mathbf{0}\), so \(\lambda = 0\) (because \(\mathbf{x} \neq \mathbf{0}\)).
Let \(A\) be diagonalizable with real eigenvalues and assume that \(A^{m} = I\) for some \(m \geq 1\).
- Show that \(A^{2} = I\).
- If \(m\) is odd, show that \(A = I\). [Hint: Theorem A.3]
- Answer
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- If \(A\mathbf{x} = \lambda\mathbf{x}\), then \(A^{k}\mathbf{x} = \lambda^{k}\mathbf{x}\) for each \(k\). Hence \(\lambda^{m}\mathbf{x} = A^{m}\mathbf{x} = \mathbf{x}\), so \(\lambda^{m} = 1\). As \(\lambda\) is real, \(\lambda = \pm 1\) by the Hint. So if \(P^{-1}AP = D\) is diagonal, then \(D^{2} = I\) by Theorem 3.4.1. Hence \(A^{2} = PD^{2}P = I\).
Let \(A^{2} = I\), and assume that \(A \neq I\) and \(A \neq -I\).
- Show that the only eigenvalues of \(A\) are \(\lambda = 1\) and \(\lambda = -1\).
- Show that \(A\) is diagonalizable. [Hint: Verify that \(A(A + I) = A + I\) and \(A(A - I) = -(A - I)\), and then look at nonzero columns of \(A + I\) and of \(A - I\).]
- If \(Q_{m} : \mathbb{R}^2 \to \mathbb{R}^2\) is reflection in the line \(y = mx\) where \(m \neq 0\), use (b) to show that the matrix of \(Q_{m}\) is diagonalizable for each \(m\).
- Now prove (c) geometrically using Theorem 3.3.3.
Let \(A^{2} = I\), and assume that \(A \neq I\) and \(A \neq -I\).
Let \(A = \left[ \begin{array}{rrr} 2 & 3 & -3 \\ 1 & 0 & -1 \\ 1 & 1 & -2 \end{array} \right]\) and \(B = \left[ \begin{array}{rrr} 0 & 1 & 0 \\ 3 & 0 & 1 \\ 2 & 0 & 0 \end{array} \right]\). Show that \(c_{A}(x) = c_{B}(x) = (x + 1)^{2} (x - 2)\), but \(A\) is diagonalizable and \(B\) is not.
- Show that the only diagonalizable matrix \(A\) that has only one eigenvalue \(\lambda\) is the scalar matrix \(A = \lambda I\).
- Is \(\left[ \begin{array}{rr} 3 & -2 \\ 2 & -1 \end{array}\right]\) diagonalizable?
- Answer
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- We have \(P^{-1}AP = \lambda I\) by the diagonalization algorithm, so \(A = P(\lambda I)P^{-1} = \lambda PP^{-1} = \lambda I\).
- No. \(\lambda = 1\) is the only eigenvalue.
Characterize the diagonalizable \(n \times n\) matrices \(A\) such that \(A^{2} - 3A + 2I = 0\) in terms of their eigenvalues. [Hint: Theorem 3.3.1.]
Let \(A = \left[ \begin{array}{cc} B & 0 \\ 0 & C \end{array}\right]\) where \(B\) and \(C\) are square matrices.
- If \(B\) and \(C\) are diagonalizable via \(Q\) and \(R\) (that is, \(Q^{-1}BQ\) and \(R^{-1}CR\) are diagonal), show that \(A\) is diagonalizable via \(\left[ \begin{array}{cc} Q & 0 \\ 0 & R \end{array}\right]\)
- Use (a) to diagonalize \(A\) if \(B = \left[ \begin{array}{rr} 5 & 3 \\ 3 & 5 \end{array}\right]\) and \(C = \left[ \begin{array}{rr} 7 & -1 \\ -1 & 7 \end{array}\right]\).
Let \(A = \left[ \begin{array}{cc} B & 0 \\ 0 & C \end{array}\right]\) where \(B\) and \(C\) are square matrices.
- Show that \(c_{A}(x) = c_{B}(x)c_{C}(x)\).
- If \(\mathbf{x}\) and \(\mathbf{y}\) are eigenvectors of \(B\) and \(C\), respectively, show that \(\left[ \begin{array}{c} \mathbf{x} \\ 0 \end{array}\right]\) and \(\left[ \begin{array}{c} 0 \\ \mathbf{y} \end{array}\right]\) are eigenvectors of \(A\), and show how every eigenvector of \(A\) arises from such eigenvectors.


