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 18E from Chapter 7.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 18E
Chapter:
Problem:
In Exercises 17–24, A is an m × n matrix with a singular value decomposition , where U is an m × m orthogonal matrix,
Step-by-Step Solution
Given Information
Given that A is square and invertible. We have to find a singular value decomposition of $A^{-1}$.
Step-1:
As the matrix V is an orthogonal matrix, $V ^ { T } = V ^ { - 1 }$, thus \[ A = U \sum V ^ { T } = U \sum V ^ { - 1 } \]
Step-2:
Therefore, the singular value decomposition of the matrix $A^{-1}$ can be obtained as: \[ \begin{aligned} A ^ { - 1 } & = \left( U \sum V ^ { - 1 } \right) ^ { - 1 } \\ & = V \sum ^ { - 1 } U ^ { - 1 } \\ & = V \sum ^ { - 1 } U ^ { T } \end{aligned} \] Therefore,
\[ A ^ { - 1 } = V \sum ^ { - 1 } U ^ { T } \]
Given that A is square and invertible. We have to find a singular value decomposition of $A^{-1}$.
Step-1:
As the matrix V is an orthogonal matrix, $V ^ { T } = V ^ { - 1 }$, thus \[ A = U \sum V ^ { T } = U \sum V ^ { - 1 } \]
Step-2:
Therefore, the singular value decomposition of the matrix $A^{-1}$ can be obtained as: \[ \begin{aligned} A ^ { - 1 } & = \left( U \sum V ^ { - 1 } \right) ^ { - 1 } \\ & = V \sum ^ { - 1 } U ^ { - 1 } \\ & = V \sum ^ { - 1 } U ^ { T } \end{aligned} \] Therefore,
\[ A ^ { - 1 } = V \sum ^ { - 1 } U ^ { T } \]