8.9E: An Application to Quadratic Forms Exercises
- Page ID
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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 a symmetric matrix \(A\) such that \(q = \mathbf{x}^{T}B\mathbf{x}\) takes the form \(q = \mathbf{x}^{T}A\mathbf{x}\).
- \(\left[ \begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array}\right]\)
- \(\left[ \begin{array}{rr} 1 & 1 \\ -1 & 2 \end{array}\right]\)
- \(\left[ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\)
- \(\left[ \begin{array}{rrr} 1 & 2 & -1 \\ 4 & 1 & 0 \\ 5 & -2 & 3 \end{array}\right]\)
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- \(A = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array}\right]\)
- \(A = \left[ \begin{array}{rrr} 1 & 3 & 2 \\ 3 & 1 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
In each case, find a change of variables that will diagonalize the quadratic form \(q\). Determine the index and \(rank \;\) of \(q\).
- \(q = x_{1}^2 + 2x_{1}x_{2} + x_{2}^2\)
- \(q = x_{1}^2 + 4x_{1}x_{2} + x_{2}^2\)
- \(q = x_{1}^2 + x_{2}^2 + x_{3}^2 - 4(x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})\)
- \(q = 7x_{1}^2 + x_{2}^2 + x_{3}^2 + 8x_{1}x_{2} + 8x_{1}x_{3} - 16x_{2}x_{3}\)
- \(q = 2(x_{1}^2 + x_{2}^2 + x_{3}^2 - x_{1}x_{2} + x_{1}x_{3} - x_{2}x_{3})\)
- \(q = 5x_{1}^2 + 8x_{2}^2 + 5x_{3}^2 - 4(x_{1}x_{2} + 2x_{1}x_{3} + x_{2}x_{3})\)
- \(q = x_{1}^2 - x_{3}^2 - 4x_{1}x_{2} + 4x_{2}x_{3}\)
- \(q = x_{1}^2 + x_{3}^2 - 2x_{1}x_{2} + 2x_{2}x_{3}\)
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- \(P = \frac{1}{\sqrt{2}}\left[ \begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right]\);
\(\mathbf{y} = \frac{1}{\sqrt{2}}\left[ \begin{array}{r} x_{1} + x_{2} \\ x_{1} - x_{2} \end{array}\right]\);
\(q = 3y_{1}^2 - y_{2}^2\); \(1\), \(2\) - \(P = \frac{1}{3}\left[ \begin{array}{rrr} 2 & 2 & -1 \\ 2 & -1 & 2 \\ -1 & 2 & 2 \end{array}\right]\);
\(\mathbf{y} = \frac{1}{3}\left[ \begin{array}{rcrcr} 2x_{1} & + & 2x_{2} & - & x_{3} \\ 2x_{1} & - & x_{2} & + & 2x_{3} \\ -x_{1} & + & 2x_{2} & + & 2x_{3} \end{array}\right]\);
\(q = 9y_{1}^2 + 9y_{2}^2 - 9y_{3}^2\); \(2\), \(3\) - \(P = \frac{1}{3}\left[ \begin{array}{rrr} -2 & 1 & 2 \\ 2 & 2 & 1 \\ 1 & -2 & 2 \end{array}\right]\);
\(\mathbf{y} = \frac{1}{3}\left[ \begin{array}{rcrcr} -2x_{1} & + & 2x_{2} & + & x_{3} \\ x_{1} & + & 2x_{2} & - & 2x_{3} \\ 2x_{1} & + & x_{2} & + & 2x_{3} \end{array}\right]\);
\(q = 9y_{1}^2 + 9y_{2}^2\); \(2\), \(2\) - \(P = \frac{1}{\sqrt{6}}\left[ \begin{array}{rrr} -\sqrt{2} & \sqrt{3} & 1 \\ \sqrt{2} & 0 & 2 \\ \sqrt{2} & \sqrt{3} & -1 \end{array}\right]\);
\(\mathbf{y} = \frac{1}{\sqrt{6}}\left[ \begin{array}{rcrcr} -\sqrt{2}x_{1} & + & \sqrt{2}x_{2} & + & \sqrt{2}x_{3} \\ \sqrt{3}x_{1} & & & + & \sqrt{3}x_{3} \\ x_{1} & + & 2x_{2} & - & x_{3} \end{array}\right]\);
\(q = 2y_{1}^2 + y_{2}^2 - y_{3}^2\); \(2\), \(3\)
- \(P = \frac{1}{\sqrt{2}}\left[ \begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right]\);
For each of the following, write the equation in terms of new variables so that it is in standard position, and identify the curve.
- \(xy = 1\) \(3x^{2} - 4xy = 2\) \(6x^{2} + 6xy - 2y^{2} = 5\) \(2x^{2} + 4xy + 5y^{2} = 1\)
- \(x_{1} = \frac{1}{\sqrt{5}}(2x - y)\), \(y_{1} = \frac{1}{\sqrt{5}}(x + 2y)\); \(4x_{1}^2 - y_{1}^2 = 2\); hyperbola
- \(x_{1} = \frac{1}{\sqrt{5}}(x + 2y)\), \(y_{1} = \frac{1}{\sqrt{5}}(2x - y)\); \(6x_{1}^2 + y_{1}^2 = 1\); ellipse
Consider the equation \(ax^{2} + bxy + cy^{2} = d\), where \(b \neq 0\). Introduce new variables \(x_{1}\) and \(y_{1}\) by rotating the axes counterclockwise through an angle \(\theta\). Show that the resulting equation has no \(x_{1}y_{1}\)-term if \(\theta\) is given by
\[\begin{aligned} & \cos2\theta = \frac{a - c}{\sqrt{b^2+(a-c)^2}} \\ & \sin2\theta = \frac{b}{\sqrt{b^2+(a-c)^2}}\end{aligned} \nonumber \]
[Hint: Use equation ([rotationEq2]) preceding Theorem 8.9.2 to get \(x\) and \(y\) in terms of \(x_{1}\) and \(y_{1}\), and substitute.]
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- Basis \(\{(i, 0, i), (1, 0, -1)\}\), dimension \(2\)
- Basis \(\{(1, 0, -2i), (0, 1, 1 - i)\}\), dimension \(2\)
Prove properties (1)–(5) preceding Example 8.9.4.
If \(A \stackrel{c}{\sim} B\) show that \(A\) is invertible if and only if \(B\) is invertible.
If \(\mathbf{x} = (x_{1}, \dots, x_{n})^{T}\) is a column of variables, \(A = A^{T}\) is \(n \times n\), \(B\) is \(1 \times n\), and \(c\) is a constant, \(\mathbf{x}^{T}A\mathbf{x} + B\mathbf{x} = c\) is called a quadratic equation in the variables \(x_{i}\).
- Put \(x_{1}^2 + 3x_{2}^2 + 3x_{3}^2 + 4x_{1}x_{2} - 4x_{1}x_{3} + 5x_{1} - 6x_{3} = 7\) in this form and find variables \(y_{1}\), \(y_{2}\), \(y_{3}\) as in (a).
- Answer
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- \(3y_{1}^2 + 5y_{2}^2 - y_{3}^2 - 3\sqrt{2}y_{1} + \frac{11}{3}\sqrt{3}y_{2} + \frac{2}{3}\sqrt{6}y_{3} = 7\)
\(y_{1} = \frac{1}{\sqrt{2}}(x_{2} + x_{3})\), \(y_{2} = \frac{1}{\sqrt{3}}(x_{1} + x_{2} - x_{3})\), \(y_{3} = \frac{1}{\sqrt{6}}(2x_{1} - x_{2} + x_{3})\)
- \(3y_{1}^2 + 5y_{2}^2 - y_{3}^2 - 3\sqrt{2}y_{1} + \frac{11}{3}\sqrt{3}y_{2} + \frac{2}{3}\sqrt{6}y_{3} = 7\)
Given a symmetric matrix \(A\), define \(q_{A}(\mathbf{x}) = \mathbf{x}^{T}A\mathbf{x}\). Show that \(B \stackrel{c}{\sim} A\) if and only if \(B\) is symmetric and there is an invertible matrix \(U\) such that \(q_{B}(\mathbf{x}) = q_{A}(U\mathbf{x})\) for all \(\mathbf{x}\). [Hint: Theorem 8.8.3.]
Let \(q(\mathbf{x}) = \mathbf{x}^{T}A\mathbf{x}\) be a quadratic form where \(A = A^{T}\).
- Show that \(q(\mathbf{x}) > 0\) for all \(\mathbf{x} \neq \mathbf{0}\), if and only if \(A\) is positive definite (all eigenvalues are positive). In this case, \(q\) is called positive definite.
- Show that new variables \(\mathbf{y}\) can be found such that \(q = \|\mathbf{y}\|^{2}\) and \(\mathbf{y} = U\mathbf{x}\) where \(U\) is upper triangular with positive diagonal entries. [Hint: Theorem 8.3.3.]
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- By Theorem 8.3.3 let \(A = U^{T}U\) where \(U\) is upper triangular with positive diagonal entries. Then \(q = \mathbf{x}^{T}(U^{T}U)\mathbf{x} = (U\mathbf{x})^{T}U\mathbf{x} = \| U\mathbf{x} \|^{2}\).
A bilinear form \(\beta\) on \(\mathbb{R}^n\) is a function that assigns to every pair \(\mathbf{x}\), \(\mathbf{y}\) of columns in \(\mathbb{R}^n\) a number \(\beta(\mathbf{x}, \mathbf{y})\) in such a way that
\[\begin{aligned} \beta(r\mathbf{x} + s\mathbf{y}, \mathbf{z}) = r\beta(\mathbf{x}, \mathbf{z}) + s\beta(\mathbf{y}, \mathbf{z}) \\ \beta(\mathbf{x}, r\mathbf{y} + s\mathbf{z}) = r\beta(\mathbf{x}, \mathbf{z}) + s\beta(\mathbf{x}, \mathbf{z})\end{aligned} \nonumber \]
for all \(\mathbf{x}\), \(\mathbf{y}\), \(\mathbf{z}\) in \(\mathbb{R}^n\) and \(r\), \(s\) in \(\mathbb{R}\). If \(\beta(\mathbf{x}, \mathbf{y}) = \beta(\mathbf{y}, \mathbf{x})\) for all \(\mathbf{x}\), \(\mathbf{y}\), \(\beta\) is called symmetric.
- If \(\beta\) is a bilinear form, show that an \(n \times n\) matrix \(A\) exists such that \(\beta(\mathbf{x}, \mathbf{y}) = \mathbf{x}^{T}A\mathbf{y}\) for all \(\mathbf{x}\), \(\mathbf{y}\).
- Show that \(A\) is uniquely determined by \(\beta\).
- Show that \(\beta\) is symmetric if and only if \(A = A^{T}\).