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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 \({\rm{u}} = {{\rm{x}}^ + }\).

Short Answer

Expert verified
  1. It is shown that \({{\rm{x}}^ + }\) is in Row \(A\).
  2. It is shown that\({{\rm{x}}^ + }\)is the solution of\(A{\rm{x}} = {\rm{b}}\).
  3. It is shown that \(\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|\), and equality holds if \({\bf{u}} = {{\rm{x}}^ + }\).

Step by step solution

01

Simplify for vector \({{\rm{x}}^ + }\)

It is given that the system\(A{\rm{x}} = {\rm{b}}\)is consistent and\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\), then the system\(A{\rm{x}} = {\rm{b}}\)can be simplified as follows:

\(\begin{array}{c}{{\rm{x}}^ + } = {A^ + }{\rm{b}}\\ = {A^ + }A{\rm{x}}\end{array}\)

As \({{\rm{x}}^ + } = {A^ + }A{\rm{x}}\), so\({{\rm{x}}^ + }\) is the orthogonal projection of \({\rm{x}}\) onto Row \(A\).

02

Simplify for product \(A{{\rm{x}}^ + }\)

Substitute\({{\rm{x}}^ + } = {A^ + }A{\rm{x}}\), into\(A{{\rm{x}}^ + }\), and simplify using\(A{\rm{x}} = {\rm{b}}\), as follows:

\(\begin{array}{c}A{{\rm{x}}^ + } = A\left( {{A^ + }A{\rm{x}}} \right)\\ = A{A^ + }A{\rm{x}}\\ = A{\rm{x}}\\ = {\rm{b}}\end{array}\)

Thus, it is shown that \({{\rm{x}}^ + }\) is the solution of \(A{\rm{x}} = {\rm{b}}\).

03

Prove \(\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|\)

Suppose the system\(A{\rm{x}} = {\rm{b}}\)issatisfied by another basis\({\rm{u}}\), such that,\(A{\rm{u}} = {\rm{b}}\). It is also known that\({{\rm{x}}^ + }\)is the orthogonal projection of \({\rm{x}}\) onto Row \(A\).

Apply the Pythagorean theorem on\({\left\| {\rm{u}} \right\|^2}\), as follows:

\(\begin{array}{c}{\left\| {\rm{u}} \right\|^2} = {\left\| {{{\rm{x}}^ + }} \right\|^2} + {\left\| {{\rm{u}} - {{\rm{x}}^ + }} \right\|^2}\\ \ge {\left\| {{{\rm{x}}^ + }} \right\|^2}\end{array}\)

So \(\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|\), and equality holds if \({\bf{u}} = {{\rm{x}}^ + }\).

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Most popular questions from this chapter

Classify the quadratic forms in Exercises 9鈥18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

15. \( - {\bf{3}}x_{\bf{1}}^{\bf{2}} - {\bf{7}}x_{\bf{2}}^{\bf{2}} - {\bf{10}}x_{\bf{3}}^{\bf{2}} - {\bf{10}}x_{\bf{4}}^{\bf{2}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{3}}} + {x_{\bf{1}}}{x_{\bf{4}}} + {\bf{6}}{x_{\bf{3}}}{x_{\bf{4}}}\)

25.Let \({\bf{T:}}{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\) be a linear transformation. Describe how to find a basis \(B\) for \({\mathbb{R}^n}\) and a basis \(C\) for \({\mathbb{R}^m}\) such that the matrix for \(T\) relative to \(B\) and \(C\) is an \(m \times n\) 鈥渄iagonal鈥 matrix.

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.
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  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.
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  17. If A is \(n \times n\), then A and \({A^T}A\) have the same singular values.

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

Classify the quadratic forms in Exercises 9-18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct P using the methods of Section 7.1.

9. \({\bf{4}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{4}}x_{\bf{2}}^{\bf{2}}\)

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