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Chapter 1: Problem 10

Suppose that \(U\) is a subspace of \(V\). What is \(U+U ?\)

Short Answer

Expert verified
The sum of U and U, denoted as U+U, is simply U, since adding any vector in U to itself or adding other pairs of vectors will result in elements that are already in U: \(U + U = U\)

Step by step solution

01

Understand the notation

The notation U+U denotes the sum of the sets of vectors U and U. In other words, we want to find the resulting set when we add any vector in U to itself.
02

Add a generic vector

Let's consider an arbitrary vector \(u \in U\). Scalar multiplication is closed under any subspace, so \(2u\) must also be in U. Since any element in U has an element that is double of it, U+U will contain all elements of U.
03

Adding other pairs of vectors

Now, let's pick any other two vectors \(v, w \in U\) and perform the vector addition \(v + w\). Since vector addition is closed under any subspace, the result \(v + w\) must be in U. This means, for any two vectors \(v\) and \(w\) in the subspace U, their sum is also an element of U.
04

Conclusion

From Step 2 and Step 3, we can conclude that adding any vector in U to itself, or adding other pairs of vectors, will result in elements that are already in U. Therefore, the sum of U and U is simply U: \[U + U = U\]

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Most popular questions from this chapter

Give an example of a nonempty subset \(U\) of \(R^{2}\) such that \(U\) is closed under addition and under taking additive inverses (meaning \(-u \in U\) whenever \(u \in U\) ), but \(U\) is not a subspace of \(\mathbf{R}^{2}.\)

Suppose \(a\) and \(b\) are real numbers, not both 0. Find real numbers \(c\) and \(d\) such that \\[ 1 /(a+b i)=c+d i \\]

Prove that the intersection of any collection of subspaces of \(V\) is a subspace of \(V\)

Prove that if \(a \in \mathbf{F}, v \in V,\) and \(a v=0,\) then \(a=0\) or \(v=0\)

Prove or give a counterexample: if \(U_{1}, U_{2}, W\) are subspaces of \(V\) such that \\[ U_{1}+W=U_{2}+W \\] \(\operatorname{then} U_{1}=U_{2}\)
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