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Chapter 1: Problem 21

Prove that a vector space is infinite-dimensional if and only if it contains an infinite linearly independent subset.

Short Answer

Expert verified
To prove that a vector space is infinite-dimensional if and only if it contains an infinite linearly independent subset, we must show two parts: 1. If a vector space is infinite-dimensional, it contains an infinite linear independent subset. This is done by building an infinite linearly independent set S in an infinite-dimensional vector space V, starting with an arbitrary non-zero vector and continuing the process indefinitely. 2. If a vector space contains an infinite linear independent subset, it is infinite-dimensional. This is shown by noting that an infinite linearly independent subset S cannot be part of any finite basis, and thus the vector space V must be infinite-dimensional. Therefore, we have shown that a vector space is infinite-dimensional if and only if it contains an infinite linearly independent subset.

Step by step solution

01

Proving if a vector space is infinite-dimensional, it contains an infinite linear independent subset

To prove this, suppose we have a vector space V that is infinite-dimensional. By being infinite-dimensional, it means that there is no finite basis for V, which means we cannot find a finite set of vectors in V that spans the space and is linearly independent. Now, let's consider building an infinite linearly independent subset of vectors, S, in V. Begin by picking an arbitrary non-zero vector, v1 鈭� V. If we add any linearly dependent vector to {v1}, it will still be linearly dependent, so we need to find another vector that is linearly independent of v1. We know that the vector v1 does not span the entire vector space V (since V is infinite-dimensional); hence there exists a vector v2 鈭� V such that v2 is not in the span of {v1}. Then, {v1, v2} is linearly independent. We can continue this process indefinitely, always finding a new vector vn+1 鈭� V such that vn+1 is not in the span of {v1, v2, ..., vn}, and the resulting set S = {v1, v2, ...} will be an infinite linearly independent subset of V. This proves the first part. Now let's move to the second part of the proof.
02

Proving if a vector space contains an infinite linear independent subset, it is infinite-dimensional

For this, let's assume we have a vector space V containing an infinite linearly independent subset, S. This infinite linearly independent subset S cannot be part of any finite basis because any finite set of vectors cannot span (or be the basis of) an infinite linearly independent set. Since S is not part of any finite basis, it implies that V is infinite-dimensional because there is no finite set of vectors that spans V. This concludes the second part of the proof. Therefore, we have proven that a vector space is infinite-dimensional if and only if it contains an infinite linearly independent subset.

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Most popular questions from this chapter

Let \(f, g, \in \mathcal{F}(R, R)\) be the functions defined by \(f(t)=e^{r t}\) and \(g(t)=e^{s t}\), where \(r \neq s\). Prove that \(f\) and \(g\) are linearly independent in \(\mathcal{F}(R, R)\).

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Let $\mathrm{V}=\left\\{\left(a_{1}, a_{2}\right): a_{1}, a_{2} \in F\right\\}\(, where \)F\( is a field. Define addition of elements of \)\mathrm{V}$ coordinatewise, and for \(c \in F\) and $\left(a_{1}, a_{2}\right) \in \mathrm{V}$, define $$ c\left(a_{1}, a_{2}\right)=\left(a_{1}, 0\right) . $$ Is \(\mathrm{V}\) a vector space over \(F\) with these operations? Justify your answer.
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