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I have returned to the definitions of span and linear independence (LI) in a finite-dimensional vector space.

There is a lemma in Axler (2015) that states a length of a linearly independent list is at most the length of a spanning list.

(2.23) "In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors."

My Question:

What is the relationship between a LI list and spanning list? What is the above lemma trying to say, if some sort of geometric intuition were to be added?

Reference: Axler, Sheldon J. $\textit{Linear Algebra Done Right}$, New York: Springer, 2015.

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  • $\begingroup$ Linearly independent lists are "minimal": no vector can be omitted without shrinking the collection of vectors that can be expressed as linear combinations of elements of the set. Spanning sets are "sufficient": they allow you to express any vector in the vector space as linear combinations of elements of the set. The above lemma is saying that "minimal sets" never have more elements than "sufficient sets". In addition to this, but not contained in that theorem, is that every linearly independent list can be extended to a spanning list, and all spanning lists can be pared down to a LI. $\endgroup$ – Arturo Magidin Aug 7 '19 at 20:56
  • $\begingroup$ @ArturoMagidin "have are"? $\endgroup$ – mathworker21 Aug 7 '19 at 20:56
  • $\begingroup$ @mathworker21: Because that was the most important part of the comment, right? $\endgroup$ – Arturo Magidin Aug 7 '19 at 21:00
  • $\begingroup$ @ArturoMagidin i dont get it. i didnt say that $\endgroup$ – mathworker21 Aug 7 '19 at 21:02
  • $\begingroup$ @mathworker21: Comments often have typos/grammar errors. Unless they radically change the meaning, there's little reason to go out of your way to point them out. Did you get it now? $\endgroup$ – Arturo Magidin Aug 7 '19 at 21:03
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One way to think about a linearly independent list of vectors is:

A list of vectors is linearly independent if and only if removing any vector from the list will result in a list whose span is strictly smaller than that of the original list.

Intuitively, the list is minimal for its span: remove any vector, you get a strictly smaller span. Intuitively, the list doesn't have any (linear redundancies).

Another, more intrinsic way of thinking about linearly independent lists is:

A list of vectors is linearly independent if and only if no vector in the list is a linear combination of the other vectors in the list.

One way to think about a spanning set for a vector space is:

A list of vectors in $V$ is a spanning set if every vector of $V$ is in the span of the list.

Intuitively, the list is "sufficient" to get you all vectors in $V$ (via linear combinations).

Note that "linearly independent" is intrinsic: it depends on the vectors (and the vector space operations), and only on them. Whereas "spanning set" is extrinsic: whether a set of vectors spans depends on which vector space you are working on. (E.g., $\{1,x,x^2\}$ is a spanning set for the vector space of real polynomials of degree at most $2$, but not for the vector space of all real polynomials.)

What the Lemma says is that spanning sets have to at least as large as linearly independent sets.

This is not trivial, and in fact turns on the fact that your scalars come from a field. To see that this assertion is not trivial, imagine that instead of a vector space where you can multiply by any element of the field, we will only take linear combinations with integer coefficients, and consider the "vector space" (in fact, it's called a module, or a $\mathbb{Z}$-module) of all integers. Here, the list consisting of $2$ and $3$ is minimal: you can get any integer with an (integral) linear combination of $2$ and $3$; but if you drop either of them, you can't get them all. You will get either just multiples of $2$, or just multiples of $3$.

On the other hand, the set consisting only of $1$ is a spanning set: every integer is an integer linear combination of $1$. So in this situation, we have a "linearly independent" set (a minimal set with respect to linear combinations) that has more elements than a spanning set.

(Caveat: There are multiple ways of defining "linearly independent", which are equivalent in a vector space; in this setting, they wouldn't be. For example, the definition that says that a list of vectors $v_1,\ldots,v_k$ is linearly independent if and only if whenever we have a linear combination equal to $0$, $\alpha_1 v_1+\cdots + \alpha_k v_k = \mathbf{0}$, all scalars must be zero: $\alpha_1=\cdots=\alpha_k=0$. Under this definition, the list I gave would not be "integrally linearly independent" because we can get $0$ as $3(2) -2(3) = 0$.)

In a vector space, the most basic relationship between linearly independent sets and spanning sets is that of the Lemma. In fact, the lemma can be refined to say that every linearly independent set can be extended to a set that is both linearly independent and spans; and every spanning set contains a spanning set that is linearly independent. From this you will show that any two linearly independent spanning sets for the same vector space have the same number of elements. That number is called the "dimension" of the vector space, and it is an invariant of fundamental importance in Linear Algebra.

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  • $\begingroup$ thank you for the response!! :) $\endgroup$ – Frank Swanton Aug 8 '19 at 21:05
  • $\begingroup$ very cool example with the $\mathbb{Z}$-module. $\endgroup$ – Frank Swanton Aug 8 '19 at 21:14
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Think about some examples in the real plane.

$$ ((1,0), (1,1)) $$ is both an independent list and a spanning list.

$$ ((1,0), (1,1), (1,2)) $$ is both a spanning list but not an independent list.

$$ ( (1,1) ) $$ is an indpendent list, but not a spanning list.

If you looked at all the independent lists you would discover that they have $0$ or $1$ or $2$ elements. You will never find one with $3$ or more elements.

If you looked at all the spanning lists you would discover that they have $2$ or $3$ or more elements. You will never find one with fewer than $2$ elements.

What Axler proves in this Lemma is that what you see in the example is typical: every spanning list is at least as long as every independent list.

The geometric intuition says that a spanning list - one whose linear combinations generate everything - must have at least as many elements as an independent list - one whose linear combinations represent what they span uniquely. The examples in the plane should support that intuition.

I don't know the book, but I suspect he will soon use this lemma to prove that all the lists that are both independent and spanning have the same length - which matches your intuition about what the dimension of the space should be.

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  • $\begingroup$ thank you for the response!! :) $\endgroup$ – Frank Swanton Aug 8 '19 at 21:05
  • $\begingroup$ nice example with a geometric intuition! $\endgroup$ – Frank Swanton Aug 8 '19 at 21:18

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