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Chapter 7: Q1E (page 395)

Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

Short Answer

Expert verified

a. For \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\)

b.For \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 185\)

c.For \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 16\)

Step by step solution

01

Find \({x^T}Ax\)

(a)

The given matrix is\(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\), and the given vector is \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\).

Find \({{\bf{x}}^T}A{\bf{x}}\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{5{x_1} + \frac{1}{3}{x_2}}\\{\frac{1}{3}{x_1} + {x_2}}\end{aligned}} \right)\\ &= {x_1}\left( {5{x_1} + \frac{1}{3}{x_2}} \right) + {x_2}\left( {\frac{1}{3}{x_1} + {x_2}} \right)\\ &= 5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2{\rm{ }}\left( 1 \right)\end{aligned}\)

Hence, the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\).

02

Find \({x^T}Ax\) for \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

(b)

As the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) when \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\).

So, for \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\), simplify \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) by substituting \({x_1} = 6,{x_2} = 1\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 5{\left( 6 \right)^2} + \left( {\frac{2}{3}} \right)\left( 6 \right)\left( 1 \right) + {\left( 1 \right)^2}\\ &= 185\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 185.

03

Find \({x^T}Ax\) for  \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

(c)

For \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\), simplify \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) by substituting \({x_1} = 1,{x_2} = 3\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 5{\left( 1 \right)^2} + \left( {\frac{2}{3}} \right)\left( 1 \right)\left( 3 \right) + {\left( 3 \right)^2}\\ &= 16\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 16.

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Most popular questions from this chapter

Question: Repeat Exercise 15 for the following SVD of a \({\bf{3 \times 4}}\) matrix \(A\):

\(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{.86}}}&{ - {\bf{.11}}}&{ - {\bf{.50}}}\\{{\bf{.31}}}&{{\bf{.68}}}&{ - {\bf{.67}}}\\{{\bf{.41}}}&{ - {\bf{.73}}}&{ - {\bf{5}}{\bf{.5}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\bf{12}}{\bf{.48}}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{{\bf{6}}{\bf{.34}}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right){\bf{ \times }}\left( {\begin{array}{*{20}{c}}{{\bf{.66}}}&{ - {\bf{.03}}}&{ - {\bf{.35}}}&{{\bf{.66}}}\\{ - {\bf{1}}{\bf{.3}}}&{ - {\bf{.90}}}&{ - {\bf{.39}}}&{ - {\bf{.13}}}\\{{\bf{.65}}}&{{\bf{.08}}}&{ - {\bf{.16}}}&{ - {\bf{.73}}}\\{ - {\bf{.34}}}&{{\bf{.42}}}&{ - {\bf{8}}{\bf{.4}}}&{ - {\bf{0}}{\bf{.8}}}\end{array}} \right)\)

Question: 12. Exercises 12鈥�14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Verify the properties of\({A^ + }\):

a. For each\({\rm{y}}\)in\({\mathbb{R}^m}\),\(A{A^ + }{\rm{y}}\)is the orthogonal projection of\({\rm{y}}\)onto\({\rm{Col}}\,A\).

b. For each\({\rm{x}}\)in\({\mathbb{R}^n}\),\({A^ + }A{\rm{x}}\)is the orthogonal projection of\({\rm{x}}\)onto\({\rm{Row}}\,A\).

c. \(A{A^ + }A = A\)and \({A^ + }A{A^ + } = {A^ + }\).

Question: Mark Each statement True or False. Justify each answer. In each part, A represents an \(n \times n\) matrix.

  1. If A is orthogonally diagonizable, then A is symmetric.
  2. If A is an orthogonal matrix, then A is symmetric.
  3. If A is an orthogonal matrix, then \(\left\| {A{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).
  4. The principal axes of a quadratic from \({{\bf{x}}^T}A{\bf{x}}\) can be the columns of any matrix P that diagonalizes A.
  5. If P is an \(n \times n\) matrix with orthogonal columns, then \({P^T} = {P^{ - {\bf{1}}}}\).
  6. If every coefficient in a quadratic form is positive, then the quadratic form is positive definite.
  7. If \({{\bf{x}}^T}A{\bf{x}} > {\bf{0}}\) for some x, then the quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is positive definite.
  8. By a suitable change of variable, any quadratic form can be changed into one with no cross-product term.
  9. The largest value of a quadratic form \({{\bf{x}}^T}A{\bf{x}}\), for \(\left\| {\bf{x}} \right\| = {\bf{1}}\) is the largest entery on the diagonal A.
  10. The maximum value of a positive definite quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is the greatest eigenvalue of A.
  11. A positive definite quadratic form can be changed into a negative definite form by a suitable change of variable \({\bf{x}} = P{\bf{u}}\), for some orthogonal matrix P.
  12. An indefinite quadratic form is one whose eigenvalues are not definite.
  13. If P is an \(n \times n\) orthogonal matrix, then the change of variable \({\bf{x}} = P{\bf{u}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form whose matrix is \({P^{ - {\bf{1}}}}AP\).
  14. If U is \(m \times n\) with orthogonal columns, then \(U{U^T}{\bf{x}}\) is the orthogonal projection of x onto ColU.
  15. If B is \(m \times n\) and x is a unit vector in \({\mathbb{R}^n}\), then \(\left\| {B{\bf{x}}} \right\| \le {\sigma _{\bf{1}}}\), where \({\sigma _{\bf{1}}}\) is the first singular value of B.
  16. A singular value decomposition of an \(m \times n\) matrix B can be written as \(B = P\Sigma Q\), where P is an \(m \times n\) orthogonal matrix and \(\Sigma \) is an \(m \times n\) diagonal matrix.
  17. If A is \(n \times n\), then A and \({A^T}A\) have the same singular values.

Determine which of the matrices in Exercises 7鈥�12 are orthogonal. If orthogonal, find the inverse.

12. \(P = \left( {\begin{aligned}{{}}{.5}&{.5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{.5}\\{.5}&{ - .5}&{ - .5}&{.5}\\{.5}&{ - .5}&{.5}&{ - .5}\end{aligned}} \right)\)

Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

37. \(\left( {\begin{aligned}{{}}{\bf{6}}&{\bf{2}}&{\bf{9}}&{ - {\bf{6}}}\\{\bf{2}}&{\bf{6}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{9}}&{ - {\bf{6}}}&{\bf{6}}&{\bf{2}}\\{\bf{6}}&{\bf{9}}&{\bf{2}}&{\bf{6}}\end{aligned}} \right)\)
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