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Chapter 7: Q5E (page 395)

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\).

a. \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\)

b. \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\)

Short Answer

Expert verified
  1. The matrix for the quadratic form \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\) is \(\left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)\).
  1. The matrix for the quadratic form \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\) is \(\left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)\).

Step by step solution

01

Matrix of the quadratic form

The coefficients of the square of variables, that is \(x_i^2\), are to be divided on the main diagonal as per the order, and the coefficient of term \({x_i}{x_j}{\rm{ }}\left( {i \ne j} \right)\), divided by 2 to split properly between the entries \(\left( {i,j} \right)\) and \(\left( {j,i} \right)\).

02

Find the corresponding matrix of \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

(a)

The given quadratic expression is \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\).

For this expression, the order of the square matrix is 3.

So, let \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\)

According to the rule in steps (1), 3, 2, and \( - 5\), will be on the main diagonal of the matrix, like \({a_{11}} = 3\), \({a_{22}} = 2\) and \({a_{33}} = - 5\).

And,\({a_{12}} = \frac{{ - 6}}{2}\), \({a_{21}} = \frac{{ - 6}}{2}\), \({a_{13}} = \frac{8}{2}\), \({a_{31}} = \frac{8}{2}\) and \({a_{23}} = \frac{{ - 4}}{2}\), \({a_{32}} = \frac{{ - 4}}{2}\).

Substitute the required value into \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\).

\(A = \left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)\)

So, the required matrix is \(\left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)\).

03

Find the corresponding matrix of \(2{x_1}{x_2}\)

(b)

The given quadratic expression is \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\), which can be written as \(0x_1^2 + 0x_2^2 + 0x_3^2 + 6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\).

For this expression, the order of the square matrix is 3.

So, let \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\)

According to the rule in step (1), 0, 0 and 0, will be on the main diagonal of the matrix, like \({a_{11}} = 0\), \({a_{22}} = 0\) and \({a_{33}} = 0\).

And \({a_{12}} = \frac{6}{2}\), \({a_{21}} = \frac{6}{2}\), \({a_{13}} = \frac{2}{2}\), \({a_{31}} = \frac{2}{2}\) and \({a_{23}} = \frac{{ - 10}}{2}\), \({a_{32}} = \frac{{ - 10}}{2}\) .

Substitute the required value into \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\).

\(A = \left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)\)

So, the required matrix is \(\left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)\).

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Most popular questions from this chapter

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17鈥�22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

14. \(\left( {\begin{aligned}{{}}{\,1}&{ - 5}\\{ - 5}&{\,\,1}\end{aligned}} \right)\)

Question: 14. 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 \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Given any \({\rm{b}}\) in \({\mathbb{R}^m}\), adapt Exercise 13 to show that \({A^ + }{\rm{b}}\) is the least-squares solution of minimum length. [Hint: Consider the equation \(A{\rm{x}} = {\rm{b}}\), where \(\mathop {\rm{b}}\limits^\^ \) is the orthogonal projection of \({\rm{b}}\) onto Col \(A\).

Determine which of the matrices in Exercises 1鈥�6 are symmetric.

3. \(\left( {\begin{aligned}{{}}2&{\,\,3}\\{\bf{2}}&4\end{aligned}} \right)\)

Question: 13. 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 \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Suppose the equation\(A{\rm{x}} = {\rm{b}}\)is consistent, and let\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\). By Exercise 23 in Section 6.3, there is exactly one vector\({\rm{p}}\)in Row\(A\)such that\(A{\rm{p}} = {\rm{b}}\). The following steps prove that\({{\rm{x}}^ + } = {\rm{p}}\)and\({{\rm{x}}^ + }\)is the minimum length solution of\(A{\rm{x}} = {\rm{b}}\).

  1. Show that \({{\rm{x}}^ + }\) is in Row \(A\). (Hint: Write \({\rm{b}}\) as \(A{\rm{x}}\) for some \({\rm{x}}\), and use Exercise 12.)
  2. Show that\({{\rm{x}}^ + }\)is a solution of\(A{\rm{x}} = {\rm{b}}\).
  3. Show that if \({\rm{u}}\) is any solution of \(A{\rm{x}} = {\rm{b}}\), then \(\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|\), with equality only if \({\rm{u}} = {{\rm{x}}^ + }\).

Question: 6. Let A be an \(n \times n\) symmetric matrix. Use Exercise 5 and an eigenvector basis for \({\mathbb{R}^n}\) to give a second proof of the decomposition in Exercise 4(b).
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