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Chapter 1: Problem 19

Let \(W_{1}\) and \(W_{2}\) be subspaces of a vector space \(V\). Prove that $W_{1} \cup W_{2}\( is a subspace of \)V\( if and only if \)W_{1} \subseteq W_{2}$ or \(W_{2} \subseteq W_{1}\).

Short Answer

Expert verified
In summary, we have shown that if the union of two subspaces W1 and W2 is a subspace of V, then either W1 is a subset of W2 or W2 is a subset of W1. Conversely, if either W1 is a subset of W2 or W2 is a subset of W1, then the union W1 鈭� W2 is a subspace of V, being closed under vector addition and scalar multiplication and containing the zero vector.

Step by step solution

01

Assume W1 鈭� W2 is a Subspace of V

To prove this direction, we will start by assuming that W1 鈭� W2 is a subspace of V. This means it must be closed under vector addition and scalar multiplication.
02

Contradict that W1 and W2 are Both Not Subsets

Now, let's prove it by contradiction. So, we'll assume that neither W1 鈯� W2 nor W2 鈯� W1. This means there must exist a vector x in W1 that is not in W2, and a vector y in W2 that is not in W1.
03

Consider the Sum x+y

Since W1 鈭� W2 is a subspace of V, the sum x+y must be in W1 鈭� W2. However, this sum cannot be in W1 since y 鈭� W1, and it cannot be in W2 since x 鈭� W2. This leads to a contradiction since every element in a subspace should be closed under addition. Therefore, either W1 鈯� W2 or W2 鈯� W1. 2. If either W1 鈯� W2 or W2 鈯� W1, then W1 鈭� W2 must be a subspace of V.
04

Assume Either W1 is a Subset of W2 or W2 is a Subset of W1

Now, let's prove the second direction. We will assume that either W1 鈯� W2 or W2 鈯� W1.
05

Prove Closure under Vector Addition

Since one of the subspaces is a subset of the other, any x 鈭� W1 and y 鈭� W2 should be either both in W1 or both in W2. Therefore, x + y 鈭� W1 鈭� W2, making W1 鈭� W2 closed under vector addition.
06

Prove Closure under Scalar Multiplication

Now, let's consider scalar multiplication. For any scalar c and any x 鈭� W1 鈭� W2, the vector cx should be in either W1 or W2 since both W1 and W2 are subspaces and are closed under scalar multiplication. Therefore, cx 鈭� W1 鈭� W2, making it closed under scalar multiplication.
07

Conclude W1 鈭� W2 is a Subspace

Since W1 鈭� W2 is closed under vector addition and scalar multiplication and contains the zero vector (as both W1 and W2 contain it), it must be a subspace of V.

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