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 8.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 25E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We have to prove that \[\operatorname { aff } ( A \cap B ) \subset ( \operatorname { aff } A \cap \operatorname { aff } B )\]
Step-1:
Let,\[\mathbf { x } , \mathbf { y } \in A \cap B\]Then, \[\mathbf { x } , \mathbf { y } \in A \text { and } \mathbf { x } , \mathbf { y } \in B\]
Step-2:
By the definition of affine set A\[( 1 - t ) \mathbf { x } + t \mathbf { y } \in \operatorname { aff } A\]By the definition of affine set B\[( 1 - t ) \mathbf { x } + t \mathbf { y } \in \operatorname { aff } B\]By combining both: \[( 1 - t ) \mathbf { x } + t \mathbf { y } \in ( \operatorname { aff } A \cap \operatorname { aff } \mathbf { B } )\] Hence, \[( 1 - t ) \mathbf { x } + t \mathbf { y } \in \operatorname { aff } ( A \cap B ) \Rightarrow ( 1 - t ) \mathbf { x } + t \mathbf { y } \in ( \operatorname { aff } A \cap \operatorname { aff } B )\]Therefore,
\[\operatorname { aff } ( A \cap B ) \subset ( \operatorname { aff } A \cap \operatorname { aff } B )\]
We have to prove that \[\operatorname { aff } ( A \cap B ) \subset ( \operatorname { aff } A \cap \operatorname { aff } B )\]
Step-1:
Let,\[\mathbf { x } , \mathbf { y } \in A \cap B\]Then, \[\mathbf { x } , \mathbf { y } \in A \text { and } \mathbf { x } , \mathbf { y } \in B\]
Step-2:
By the definition of affine set A\[( 1 - t ) \mathbf { x } + t \mathbf { y } \in \operatorname { aff } A\]By the definition of affine set B\[( 1 - t ) \mathbf { x } + t \mathbf { y } \in \operatorname { aff } B\]By combining both: \[( 1 - t ) \mathbf { x } + t \mathbf { y } \in ( \operatorname { aff } A \cap \operatorname { aff } \mathbf { B } )\] Hence, \[( 1 - t ) \mathbf { x } + t \mathbf { y } \in \operatorname { aff } ( A \cap B ) \Rightarrow ( 1 - t ) \mathbf { x } + t \mathbf { y } \in ( \operatorname { aff } A \cap \operatorname { aff } B )\]Therefore,
\[\operatorname { aff } ( A \cap B ) \subset ( \operatorname { aff } A \cap \operatorname { aff } B )\]