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The first chapter began by introducing Gauss' method and finished with a fair understanding, keyed on the Linear Combination Lemma, of how it finds the solution set of a linear system. Gauss' method systematically takes linear combinations of the rows. With that insight, we now move to a general study of linear combinations.

We need a setting for this study. At times in the first chapter, we've combined vectors from {ℝ}^{2} , at other times vectors from {ℝ}^{3} , and at other times vectors from even higher-dimensional spaces. Thus, our first impulse might be to work in {ℝ}^{n} , leaving n unspecified. This would have the advantage that any of the results would hold for {ℝ}^{2} and for {ℝ}^{3} and for many other spaces, simultaneously.

But, if having the results apply to many spaces at once is advantageous then sticking only to {ℝ}^{n}'s is overly restrictive. We'd like the results to also apply to combinations of row vectors, as in the final section of the first chapter. We've even seen some spaces that are not just a collection of all of the same-sized column vectors or row vectors. For instance, we've seen a solution set of a homogeneous system that is a plane, inside of {ℝ}^{3} . This solution set is a closed system in the sense that a linear combination of these solutions is also a solution. But it is not just a collection of all of the three-tall column vectors; only some of them are in this solution set.

We want the results about linear combinations to apply anywhere that linear combinations are sensible. We shall call any such set a vector space. Our results, instead of being phrased as "Whenever we have a collection in which we can sensibly take linear combinations ...", will be stated as "In any vector space ...".

Such a statement describes at once what happens in many spaces. The step up in abstraction from studying a single space at a time to studying a class of spaces can be hard to make. To understand its advantages, consider this analogy. Imagine that the government made laws one person at a time: "Leslie Jones can't jay walk." That would be a bad idea; statements have the virtue of economy when they apply to many cases at once. Or, suppose that they ruled, "Kim Ke must stop when passing the scene of an accident." Contrast that with, "Any doctor must stop when passing the scene of an accident." More general statements, in some ways, are clearer.

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Next in Linear Algebra

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Definition 1.1 A **vector space** (over $ℝ$) consists of a set V along with two operations "$+$" and "$⋅$" subject to these conditions. - For any $\overset{→}{v} , \overset{→}{w} ∈ V : \overset{→}{v} + \overset{→}{w} ∈ V$ . - For any $\overset{→}{v} , \overset{→}{w} ∈ V : \overset{→}{v} + \overset{→
CC BY-SA 4.0
One of the examples that led us to introduce the idea of a vector space was the solution set of a homogeneous system. For instance, we've seen in [Example 1.4]( such a space that is a planar subset of ${ℝ}^{3}$. There, the vector space ${ℝ}^{3}$ contains inside it another vector space, the plane. De
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The prior section shows that a vector space can be understood as an unrestricted linear combination of some of its elements— that is, as a span. For example, the space of linear polynomials $\{ a + b x \ | \ a , b ∈ ℝ \}$ is spanned by the set $\{ 1 , x \}$. The prior section also showed that a spac
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For the right to left implication assume that $\overset{→}{v} ∈ [ S ]$ to show that $[ S ] = [ S ∪ \{ \overset{→}{v} \} ]$ by mutual inclusion. The inclusion $[ S ] ⊆ [ S ∪ \{ \overset{→}{v} \} ]$ is obvious. For the other inclusion $[ S ] ⊇ [ S ∪ \{ \overset{→}{v} \} ]$, write an element of $[ S ∪
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Definition 1.1 A **basis** for a vector space is a sequence of vectors that form a set that is linearly independent and that spans the space. We denote a basis with angle brackets $⟨ \overset{→}{{β}_{1}} , \overset{→}{{β}_{2}} , … ⟩$ to signify that this collection is a sequence[1] — the order of th
CC BY-SA 4.0
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