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Chapter 4: Problem 48

Complete the proof of Theorem 4.6 by showing that the intersection of two subspaces of a vector space is closed under scalar multiplication.

Short Answer

Expert verified
The intersection of two subspaces, \( V_1 \) and \( V_2 \), denoted as \( V = V_1 \cap V_2 \) is closed under scalar multiplication. For every vector \( v \) in \( V \), if we multiply it with a scalar \( c \), the resulting vector \( cv \) will also be in \( V_1 \) and \( V_2 \) by rules of scalar multiplication in subspaces, and hence, \( cv \) is also in the intersection \( V \). So, the intersection \( V \) is closed under scalar multiplication.

Step by step solution

01

Define the Subspaces

We'll denote the two vector subspaces as \( V_1 \) and \( V_2 \), and their intersection as \( V = V_1 \cap V_2 \). For any vector \( v \) in \( V \), \( v \) is also in \( V_1 \) and \( V_2 \) since \( V \) is the intersection of \( V_1 \) and \( V_2 \).
02

Scalar Multiplication in Individual Subspaces

In vector space subspaces, any vector, when multiplied by a scalar should result in another vector within the same subspace. Therefore, if we have a scalar \( c \) and we multiply \( c \) with \( v \), \( cv \) is in \( V_1 \) and \( V_2 \), since both \( V_1 \) and \( V_2 \) are subspaces.
03

Scalar Multiplication in the Intersection of Subspaces

Since \( cv \) is in both \( V_1 \) and \( V_2 \), it is also in their intersection, \( V \). Thus \( V \) is closed under scalar multiplication. This proves that the intersection of any two subspaces of a vector space is also closed under scalar multiplication.

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