Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 25E from Chapter 2.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 25E
Chapter:
Problem:
Exercises 22–26 provide a glimpse of some widely used matrix factorizations, some of which are discussed later in the text.
Step-by-Step Solution
Given Information
Given that U and V are $n \times n$ matrices such that $U ^ { \top } U = I$ and $V ^ { \top } V = I$ and \[ A = U D V ^ { T } \] We have to Show that A is invertible, and find a formula for the same
Step-1:
To prove that U is invertible, we have to show that determinant of U is invertible. \[ \begin{aligned} U ^ { \top } U & = I \\ \operatorname { det } \left( U ^ { \top } U \right) & = \operatorname { det } ( I ) \\ \operatorname { det } ( U ) \operatorname { det } \left( U ^ { \tau } \right) & = 1 \\ \operatorname { det } ( U ) \operatorname { det } ( U ) & = 1 \\ [ \operatorname { det } ( U ) ] ^ { 2 } & = 1 \\ \operatorname { det } ( U ) & = \pm 1 \end{aligned} \] Therefore,
The matrix U is invertible
Step-2:
Calculate the inverse of matrix U as follows: \[ \begin{aligned} \left( U ^ { \top } U \right) U ^ { - 1 } & = U ^ { - 1 } \\ U ^ { \mathrm { T } } \left( U U ^ { - 1 } \right) & = U ^ { - 1 } \\ U ^ { \mathrm { T } } ( I ) & = U ^ { - 1 } \\ U ^ { \mathrm { T } } & = U ^ { - 1 } \end{aligned} \] Therefore,
The inverse of matrix U is $U^T$
Step-3:
Show that determinant of V is nonzero \[ \begin{aligned} V ^ { \mathrm { T } } V & = I \\ \operatorname { det } \left( V ^ { \mathrm { T } } V \right) & = \operatorname { det } ( I ) \\ \operatorname { det } ( V ) \operatorname { det } \left( V ^ { \tau } \right) & = 1 \\ \operatorname { det } ( V ) \operatorname { det } ( V ) & = 1 \\ [ \operatorname { det } ( V ) ] ^ { 2 } & = 1 \\ \operatorname { det } ( V ) & = \pm 1 \end{aligned} \] Therefore,
The matrix V is invertible
Step-4:
Inverse of the matrix V. \[ \begin{aligned} \left( V ^ { \mathrm { T } } V \right) V ^ { - 1 } & = V ^ { - 1 } \\ V ^ { \top } \left( V V ^ { - 1 } \right) & = V ^ { - 1 } \\ V ^ { \mathrm { T } } ( I ) & = V ^ { - 1 } \\ V ^ { \mathrm { T } } & = V ^ { - 1 } \end{aligned} \] Therefore,
The inverse of matrix V is $V^T$
Step-5: Inverse of A
\[ \begin{aligned} A ^ { - 1 } & = \left( U D V ^ { \top } \right) ^ { - 1 } \\ & = \left( U \left( D V ^ { \top } \right) \right) ^ { - 1 } \\ & = \left( D V ^ { \top } \right) ^ { - 1 } U ^ { - 1 } \\ & = \left\{ \left( V ^ { \top } \right) ^ { - 1 } D ^ { - 1 } \right\} U ^ { - 1 } \\ = & \left( V ^ { - 1 } \right) ^ { \top } D ^ { - 1 } U ^ { - 1 } \end{aligned} \] Upon substitution: \[ \begin{aligned} A ^ { - 1 } & = \left( V ^ { - 1 } \right) ^ { \mathrm { T } } D ^ { - 1 } U ^ { - 1 } \\ & = \left( V ^ { \tau } \right) ^ { \mathrm { T } } D ^ { - 1 } U ^ { T } \\ & = V D ^ { - 1 } U ^ { T } \end{aligned} \] Therefore,
The inverse of matrix A is \[{A^{ - 1}} = V{D^{ - 1}}{U^T}\]
Given that U and V are $n \times n$ matrices such that $U ^ { \top } U = I$ and $V ^ { \top } V = I$ and \[ A = U D V ^ { T } \] We have to Show that A is invertible, and find a formula for the same
Step-1:
To prove that U is invertible, we have to show that determinant of U is invertible. \[ \begin{aligned} U ^ { \top } U & = I \\ \operatorname { det } \left( U ^ { \top } U \right) & = \operatorname { det } ( I ) \\ \operatorname { det } ( U ) \operatorname { det } \left( U ^ { \tau } \right) & = 1 \\ \operatorname { det } ( U ) \operatorname { det } ( U ) & = 1 \\ [ \operatorname { det } ( U ) ] ^ { 2 } & = 1 \\ \operatorname { det } ( U ) & = \pm 1 \end{aligned} \] Therefore,
The matrix U is invertible
Step-2:
Calculate the inverse of matrix U as follows: \[ \begin{aligned} \left( U ^ { \top } U \right) U ^ { - 1 } & = U ^ { - 1 } \\ U ^ { \mathrm { T } } \left( U U ^ { - 1 } \right) & = U ^ { - 1 } \\ U ^ { \mathrm { T } } ( I ) & = U ^ { - 1 } \\ U ^ { \mathrm { T } } & = U ^ { - 1 } \end{aligned} \] Therefore,
The inverse of matrix U is $U^T$
Step-3:
Show that determinant of V is nonzero \[ \begin{aligned} V ^ { \mathrm { T } } V & = I \\ \operatorname { det } \left( V ^ { \mathrm { T } } V \right) & = \operatorname { det } ( I ) \\ \operatorname { det } ( V ) \operatorname { det } \left( V ^ { \tau } \right) & = 1 \\ \operatorname { det } ( V ) \operatorname { det } ( V ) & = 1 \\ [ \operatorname { det } ( V ) ] ^ { 2 } & = 1 \\ \operatorname { det } ( V ) & = \pm 1 \end{aligned} \] Therefore,
The matrix V is invertible
Step-4:
Inverse of the matrix V. \[ \begin{aligned} \left( V ^ { \mathrm { T } } V \right) V ^ { - 1 } & = V ^ { - 1 } \\ V ^ { \top } \left( V V ^ { - 1 } \right) & = V ^ { - 1 } \\ V ^ { \mathrm { T } } ( I ) & = V ^ { - 1 } \\ V ^ { \mathrm { T } } & = V ^ { - 1 } \end{aligned} \] Therefore,
The inverse of matrix V is $V^T$
Step-5: Inverse of A
\[ \begin{aligned} A ^ { - 1 } & = \left( U D V ^ { \top } \right) ^ { - 1 } \\ & = \left( U \left( D V ^ { \top } \right) \right) ^ { - 1 } \\ & = \left( D V ^ { \top } \right) ^ { - 1 } U ^ { - 1 } \\ & = \left\{ \left( V ^ { \top } \right) ^ { - 1 } D ^ { - 1 } \right\} U ^ { - 1 } \\ = & \left( V ^ { - 1 } \right) ^ { \top } D ^ { - 1 } U ^ { - 1 } \end{aligned} \] Upon substitution: \[ \begin{aligned} A ^ { - 1 } & = \left( V ^ { - 1 } \right) ^ { \mathrm { T } } D ^ { - 1 } U ^ { - 1 } \\ & = \left( V ^ { \tau } \right) ^ { \mathrm { T } } D ^ { - 1 } U ^ { T } \\ & = V D ^ { - 1 } U ^ { T } \end{aligned} \] Therefore,
The inverse of matrix A is \[{A^{ - 1}} = V{D^{ - 1}}{U^T}\]