4.8.E: Exercise for Section 4.8
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3} \\ \frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3}\end{array}\right],\: \left[\begin{array}{c}\frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2}\end{array}\right],\: \left[\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.
Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\2\\-1\end{array}\right],\:\left[\begin{array}{r}1\\0\\1\end{array}\right],\:\left[\begin{array}{r}-1\\1\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.
- Answer
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The given set of vectors: \[ \vec{v}_1 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \nonumber\] is orthogonal since their pairwise dot products are zero. To normalize them, we compute their norms: \[ \|\vec{v}_1\| = \sqrt{6}, \quad \|\vec{v}_2\| = \sqrt{2}, \quad \|\vec{v}_3\| = \sqrt{3} \nonumber\] Thus, an orthonormal basis with the same span is: \[ \vec{u}_1 = \frac{\vec{v}_1}{\|\vec{v}_1\|} = \begin{bmatrix} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{6}} \end{bmatrix} \nonumber\] \[ \vec{u}_2 = \frac{\vec{v}_2}{\|\vec{v}_2\|} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix} \nonumber\] \[ \vec{u}_3 = \frac{\vec{v}_3}{\|\vec{v}_3\|} = \begin{bmatrix} -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{bmatrix} \nonumber\]
Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\1\\-1\end{array}\right],\:\left[\begin{array}{r}0\\1\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.
Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\1\\-1\end{array}\right],\:\left[\begin{array}{r}1\\2\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.
Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\0\\0\\0\end{array}\right],\:\left[\begin{array}{r}0\\1\\-1\\0\end{array}\right],\:\left[\begin{array}{r}0\\0\\0\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.
Here are some matrices. Label according to whether they are symmetric, skew symmetric, or orthogonal.
- \(\left[\begin{array}{ccc}1&0&0 \\ 0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right]\)
- \(\left[\begin{array}{ccc}1&2&-3 \\ 2&1&4 \\ -3&4&7\end{array}\right]\)
- \(\left[\begin{array}{ccc}0&-2&-3 \\ 2&0&-4 \\ 3&4&0\end{array}\right]\)
- Answer
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- Orthogonal
- Symmetric
- Skew Symmetric
In this exercise, you will show that orthogonal matrices are exactly those which preserve length.
- For \(U\) an orthogonal matrix, explain why \(||U\vec{x}|| =||\vec{x}||\) for any vector \(\vec{x}\).
- Next explain why if \(U\) is an \(n\times n\) matrix with the property that \(||U\vec{x}|| =||\vec{x}||\) for all vectors, \(\vec{x}\), then \(U\) must be orthogonal.
- Answer
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- Suppose \(U\) is an orthogonal matrix. \(||U\vec{x}||^2=U\vec{x}\bullet U\vec{x}=U^TU\vec{x}\bullet\vec{x}=I\vec{x}\bullet\vec{x}=||\vec{x}||^2\).
- Now, suppose distance is preserved by \(U\). Then \[\begin{aligned} (U(\vec{x}+\vec{y}))\bullet (U(\vec{x}+\vec{y}))&=||Ux||^2+||Uy||^2+2(Ux\bullet Uy) \\ &=||\vec{x}||^2+||\vec{y}||^2+2(U^TU\vec{x}\bullet\vec{y})\end{aligned}\] But since \(U\) preserves distances, it is also the case that \[(U(\vec{x}+\vec{y})\bullet U(\vec{x}+\vec{y}))=||\vec{x}||^2+||\vec{y}||^2+2(\vec{x}\bullet\vec{y})\nonumber\] Hence \[\vec{x}\bullet\vec{y}=U^TU\vec{x}\bullet\vec{y}\nonumber\] and so \[((U^TU-I)\vec{x})\bullet\vec{y}=0\nonumber\] Since \(\vec y\) is arbitrary, it follows that \(U^TU-I=0\). Thus \(U\) is orthogonal.
Suppose \(U\) is an orthogonal \(n\times n\) matrix. Explain why \(rank(U) = n\).
- Answer
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You could observe that \(\text{det}(UU^T)=(\text{det}(U))^2-1\) so \(\text{det}(U)\neq 0\).
Fill in the missing entries to make the matrix orthogonal. \[\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\underline{\;}&\underline{\;} \\ \underline{\;}&\frac{\sqrt{6}}{3}&\underline{\;}\end{array}\right].\nonumber\]
- Answer
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\[\begin{aligned} &\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a \\ 0&\frac{\sqrt{6}}{3}&b\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a \\ 0&\frac{\sqrt{6}}{3}&b\end{array}\right]^T \\ =&\left[\begin{array}{ccc} 1&\frac{1}{3}\sqrt{3}a-\frac{1}{3} &\frac{1}{3}\sqrt{3}b-\frac{1}{3} \\ \frac{1}{3}\sqrt{3}a-\frac{1}{3}&a^2+\frac{2}{3}&ab-\frac{1}{3} \\ \frac{1}{3}\sqrt{3}b-\frac{1}{3}&ab-\frac{1}{3}&b^2+\frac{2}{3}\end{array}\right]\end{aligned}\] This requires, \(a=1/\sqrt{3},b=1/\sqrt{3}\). \[\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3} \\ 0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3} \\ 0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right]^T =\left[\begin{array}{ccc}1&0&0 \\ 0&1&0 \\ 0&0&1\end{array}\right]\nonumber\]
Fill in the missing entries to make the matrix orthogonal. \[\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\underline{\;}&\underline{\;} \\ \underline{\;}&0&\underline{\;}\end{array}\right]\nonumber\]
- Answer
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\[\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&a \\ -\frac{1}{3}&0&b\end{array}\right]\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&a \\ -\frac{1}{3}&0&b\end{array}\right]^T=\left[\begin{array}{ccc}1&\frac{1}{6}\sqrt{2}a-\frac{1}{18}&\frac{1}{6}\sqrt{2}b-\frac{2}{9} \\ \frac{1}{6}\sqrt{2}a-\frac{1}{18}&a^2+\frac{17}{18} &ab-\frac{2}{9} \\ \frac{1}{6}\sqrt{2}b-\frac{2}{9}&ab-\frac{2}{9}&b^2+\frac{1}{9}\end{array}\right]\nonumber\] This requires \(a=\frac{1}{3\sqrt{2}},\:b=\frac{4}{3\sqrt{2}}\). \[\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}} \\ -\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}} \\ -\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right]^T=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\nonumber\]
Fill in the missing entries to make the matrix orthogonal. \[\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\underline{\;} \\ \frac{2}{3}&0&\underline{\;} \\ \underline{\;}&\underline{\;}&\frac{4}{15}\sqrt{5}\end{array}\right]\nonumber\]
- Answer
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Try \[\begin{aligned}&\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c \\ \frac{2}{3}&0&d \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c \\ \frac{2}{3}&0&d \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]^T \\ =&\left[\begin{array}{ccc}c^2+\frac{41}{45} &cd+\frac{2}{9}&\frac{4}{15}\sqrt{5}c-\frac{8}{45} \\ cd+\frac{2}{9}&d^2+\frac{4}{9} &\frac{4}{15}\sqrt{5}d+\frac{4}{9} \\ \frac{4}{15}\sqrt{5}c-\frac{8}{45}&\frac{4}{15}\sqrt{5}d+\frac{4}{9}&1\end{array}\right]\end{aligned}\] This requires that \(c=\frac{2}{3\sqrt{5}},d=\frac{-5}{3\sqrt{5}}\). \[\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}} \\ \frac{2}{3}&0&\frac{-5}{3\sqrt{5}} \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}} \\ \frac{2}{3}&0&\frac{-5}{3\sqrt{5}} \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]^T=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\nonumber\]
Consider the set of \( \mathbb{R}^3 \) vectors: \[ \vec{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}. \nonumber \]
- Show that this set forms an \(\textbf{orthogonal basis}\) for \( \mathbb{R}^3 \).
- Given the vector \( \vec{w} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \nonumber \), find the Fourier expansion of \( \vec{w} \) in terms of this basis.
Consider the set of \( \mathbb{R}^4 \) vectors: \[ \vec{u}_1 = \frac{1}{2} \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \quad \vec{u}_2 = \frac{1}{2} \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix}, \quad \vec{u}_3 = \frac{1}{2} \begin{bmatrix} 1 \\ 1 \\ -1 \\ -1 \end{bmatrix}, \quad \vec{u}_4 = \frac{1}{2} \begin{bmatrix} 1 \\ -1 \\ -1 \\ 1 \end{bmatrix}. \nonumber \]
- Show that this set forms an \(\textbf{orthonormal basis}\) for \( \mathbb{R}^4 \).
- Given the vector: \( \vec{w} = \begin{bmatrix} 3 \\ 1 \\ -2 \\ 0 \end{bmatrix} \nonumber \), find the Fourier expansion of \( \vec{w} \) in terms of this basis.
Show that if a matrix \(A\) has orthogonal columns, then \(A^T A\) is a diagonal matrix.
- Hint
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Compute the \((i,j)\) entry of \(A^T A\) explicitly.
Prove that if \( Q \) is an orthogonal matrix, then it preserves inner products, i.e., for any vectors \( \vec{x}, \vec{y} \in \mathbb{R}^n \), \[ \langle Q\vec{x}, Q\vec{y} \rangle = \langle \vec{x}, \vec{y} \rangle. \nonumber\]
- Hint
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Express the inner product using matrix multiplication and use the fact that \( Q^T Q = I \).