- #1

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**d**is a vector in the vector space ℝ

^{2}, then is:

**{td | t ∈ ℝ}**the same as

**span{d}**?

Thank you.

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- Thread starter Raymondyhq
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- #1

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Thank you.

- #2

mathwonk

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yes. but you should look at a definition of "span" and check this yourself.

- #3

fresh_42

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Basically, yes. It would generally be better to speak of a linear span instead of just span, but this is a common sloppiness. Span cannot be recommended. Also in such a general context like here, it would be better to add the scalar field as an index ##\operatorname{lin}_\mathbb{R}\{d\}=\operatorname{span}_\mathbb{R}\{d\}##. Span as operatorname is no protected abbreviation, because it is context sensitive. So with even less effort one can write ##\mathbb{R}\cdot d## or ##\sum_{i\in I}\mathbb{R}d_i## if more than one vector is involved.dis a vector in the vector space ℝ^{2}, then is:

{td | t ∈ ℝ}the same asspan{d}?

Thank you.

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- #5

Mark44

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I don't think it's necessary to include "linear" with "span" since span is already defined in most textbooks to be the set of all linear combinations of a set of vectors. Also, unless we're talking about a completely arbitrary vector space, with neither the dimension nor underlying field known, we usually have some idea about the dimension of the vector space and the field from which the scalars are drawn.Basically, yes. It would generally be better to speak of a linear span instead of just span, but this is a common sloppiness.

In the example of the OP in this thread, the scalar field is clearly R.Also in such a general context like here, it would be better to add the scalar field as an index

Let us assume thatdis a vector in the vector space ℝ^{2}, then is:

{td | t ∈ ℝ}the same asspan{d}?

- #6

fresh_42

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Maybe, but it takes not much to be precise. Especially the scalar field is important. Outside physics it is not automatically clear that it is of characteristic zero or algebraically closed. And even in physics, there is a major difference between ##\mathbb{R}## and ##\mathbb{C}##. E.g. someone recently asked about the span of ##\{z,\bar{z}\}##, in which case you get two different spaces, depending on whether it is a real or complex vector space. It is correct that span is usually the linear span and others are calledSpan(d) is unambiguous in my opinion.

- #7

Mark44

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To me, writing "linear span" instead of "span" seems redundant. Two of the linear algebra books I pulled from my shelf talk only about "span" and don't further qualify it by adding "linear."Maybe, but it takes not much to be precise.

If the context is crystal clear, I don't see any advantage in being over-precise.

Sure, but the way the question was written, it was obvious what the scalar field was.Especially the scalar field is important.

Of course. And when that context is very clear, I don't see any lack of clarity by omitting redundant details. There are some men who alway use a belt and suspenders to hold their pants up.In the given example, of course, the question: Is ##\mathbb{R}d=\operatorname{span}\{d\}##? only allows a real vector space, but that is what I wanted to say:spanis context sensitive.

- #8

Mark44

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Just to be clear, the OP has identified the vector space: ##\mathbb {R^2}## and the scalar field: ##\mathbb R##. I would be willing to bet that the response of ##\mathbb R d = span\{d\}## would be confusing to the OP, rather than clarifying.Let us assume thatdis a vector in the vector space ℝ^{2}, then is:

{td | t ∈ ℝ}the same asspan{d}?

- #9

WWGD

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I don't think it's necessary to include "linear" with "span" since span is already defined in most textbooks to be the set of all linear combinations of a set of vectors. .

I have always found the equivalent definition that for any set S of vectors, Span(S) is the smallest vector space containing the set S -- the largest of course, being the whole ambient space, which will happen if S contains a basis for S.

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