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Chapter 2: Q2.5-25Q (page 93)

25. (Singular Value Decomposition) Suppose \[A = UD{V^T}\], where U and Vare \[n \times n\] matrices with the property that \[{U^T}U = I\] and \[{V^T}V = I\], and where D is a diagonal matrix with positive numbers \[{\sigma _1}, \ldots ,{\sigma _n}\] on the diagonal. Show that A is invertible, and find a formula for \[{A^{ - {\bf{1}}}}\].

Short Answer

Expert verified

Here,Ais the product of invertible matrices. Thus, Ais invertible, and the formula for \[{A^{ - 1}}\] is \[{A^{ - 1}} = V{D^{ - 1}}{U^T}\].

Step by step solution

01

Use the fact of an invertible matrix

Given, \[A = UD{V^T}\] with \[{U^T}U = I\] and \[{V^T}V = I\].

This implies that U and \[{V^T}\] are invertible matrices with \[{U^{ - 1}} = {U^T}\], and \[{\left( {{V^T}} \right)^{ - 1}} = V\].

Note that the determinant of a diagonal matrix is the product of its diagonal entries, i.e., \[\det \,D = {\sigma _1} \ldots {\sigma _n} \ne 0\]. Hence, D is also invertible.

So, here,Ais the product of invertible matrices. Thus, Ais invertible.

02

Find the inverse matrix

\[\begin{array}{c}{A^{ - 1}} = {\left( {UD{V^T}} \right)^{ - 1}}\\ = {\left( {{v^T}} \right)^{ - 1}}{\left( {UD} \right)^{ - 1}}\\{A^{ - 1}} = {\left( {{v^T}} \right)^{ - 1}}{D^{ - 1}}{U^{ - 1}}\end{array}\]

03

Find the formula for the inverse of A

Here, \[{U^{ - 1}} = {U^T}\], and \[{\left( {{V^T}} \right)^{ - 1}} = V\]. Therefore, \[{A^{ - 1}} = V{D^{ - 1}}{U^T}\].

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Most popular questions from this chapter

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

Suppose Ais an \(n \times n\) matrix with the property that the equation \[A{\mathop{\rm x}\nolimits} = 0\] has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

c. \(AB + AC = A\left( {B + C} \right)\)

d. \({A^T} + {B^T} = {\left( {A + B} \right)^T}\)

e. The transpose of a product of matrices equals the product of their transposes in the same order.

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).
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