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My question concerns the definition of linear combinations and a criteria for linear independence of a Set (either finite or infinite).

Here is the following Definition and Criteria given:

A vector $v$ in vector space $V$ is a linear combination of vectors of set $S$ if there is a a finite number of vectors $x_1, \cdots , x_n$ and scalars $a_1, \cdots , a_n$ such that $$v= a_1x_1 + \cdots + a_nx_n$$

Question:

Here is finite restricted to more than $1$ vector. Or does this definition include the possibility of a vector $v$ being a linear combination of no vectors?

Also, can these finite vectors in set S be the same. That is must they all be distinct?

Criteria for Linear Independence:

My book writes the following fact:

A set is linearly independent if and only if the only representations of the zero vector as linear combinations of its vectors are trivial representations.

Question:

Taking in mind the definition of linear dependence requires a nontrivial representation of the zero vector as a linear combination of DISTINCT vectors of the set examined for linear dependence, why does this criteria for linear Independence say "as linear combinations of its vectors" as opposed to "as linear combinations of its distinct vectors" ?

Lastly, if a set is linear independent, does that imply the set contains distinct vectors?

Does it make sense to talk about linear dependence and linear independence in the context of a set of vectors with some vectors repeated. That is with a set of vectors with element all not distinct?

Thanks in Advance.

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  • $\begingroup$ To your concern of distinctiveness, since you're talking about "set", so vectors in a set are distinct by default. $\endgroup$ – Postal Model Jul 22 '18 at 8:34
  • $\begingroup$ I'm reading the same book, my question may be helpful to you: math.stackexchange.com/q/2675139/390226 $\endgroup$ – Postal Model Jul 22 '18 at 8:38
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Here are answers to your questions:-

  1. Firstly, when you say scalars, $a_1, a_2, \cdots, a_n$, they are real numbers and hence can also be $0$. Keeping this in mind, suppose there is a set $S = \left\lbrace v_1, v_2, \cdots, v_n \right\rbrace \subseteq V$. Then, quite obviously, the vector $\textbf{0} \in V$ cab ve written as $$0 \cdot v_1 + 0 \cdot v_2 + \cdots + 0 \cdot v_n = \textbf{0}$$ This is what we call the "trivial linear combination".

In fact, what confusion you have in mind is that when you say that a vector is a linear combination of other vectors, there must be at least one vector and one scalar with which you can construct your "linear combination".

  1. When you talk about a "set", elements cannot be repeated. So, there is no point of asking if the elements of the set are distinct.

Lastly, I do not know what book you are following, but I feel that a better version of definitions of linear dependence and independence is the following:-

Linear Independence

A finite set $S = \left\lbrace v_1, v_2, \cdots, v_n \right\rbrace \subseteq V$ is said to be linearly independent iff

$$\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \cdots + \alpha_n \cdot v_n = \textbf{0}$$

implies that $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$. This actually means that the only way you can obtain the zero vector $\textbf{0}$ from a linearly "independent" set is by setting the scalars (coefficients) to be $0$, which we call the "trivial" combination.

In case of an infinite set $S \subseteq V$, it is said to be linearly independent iff every finite subset of $S$ is linearly independent. We have definition of linear independence of finite sets which can be used.

Linear Dependence

A finite set $S = \left\lbrace v_1, v_2, \cdots, v_n \right\rbrace \subseteq V$ is said to be linearly "dependent" iff it is not linearly independent. Thus, we need to negate the statement for linear independence. The negation of the statement

"$\exists \alpha_1, \alpha_2, \cdots, \alpha_n \in \mathbb{R}$ and $i \in \left\lbrace 1, 2, \cdots, n \right\rbrace$ such that $\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \cdots + \alpha_n \cdot v_n = \textbf{0}$ and $\alpha_i \neq 0$"

This statement means that the vector $v_i \in S$ can be actually written as a linear combination of the other vectors. In particular,

$$v_i = \left( - \dfrac{\alpha_1}{\alpha_i} \right) \cdot v_1 + \left( - \dfrac{\alpha_2}{\alpha_i} \right) \cdot v_2 + \cdots + \left( - \dfrac{\alpha_{i - 1}}{\alpha_i} \right) \cdot v_{i - 1} + \left( - \dfrac{\alpha_{i + 1}}{\alpha_i} \right) \cdot v_{i + 1} + \cdots + \left( - \dfrac{\alpha_n}{\alpha_i} \right) \cdot v_n$$

and therefore the vector $v_i \in S$ is "dependent" on the other vectors.

In fact, the linear combination $\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \cdots + \alpha_n \cdot v_n = \textbf{0}$ is called the "non - trivial" linear combination.

For an infinite set $S \subseteq V$, it is said to be linearly dependent iff it is not linearly independent. Again, we need to negate the statement for linear independence of infinite set. The negation of the statement would be

"There exists a finite set $A \subset S$ such that $A$ is not linearly independent". And now, we do have the definition of linear dependence (not linear independence) for finite sets which can be used.

I hope your confusion about distinct elements will be cleared by this. And if you are still confused, try forming sets which are linearly dependent and independent in $\mathbb{R}^2$ and $\mathbb{R}^3$ which you can easily visualize. Also read some material on span of a set and how we can connect linear combination and span with linear dependence and independence.

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  • $\begingroup$ Thanks for the insights. I am using linear algebra by Friedberg . It defines linear dependence of a set as a non trivial linear combination of distinct vectors in the set. Then defines linear independence as the negation of it. However, I'll stick to your given definition as it makes more sense this way. $\endgroup$ – Gabe Jul 22 '18 at 5:56
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The empty set is a finite set by all standards, so this definition also allows for taking $n=0$. That choice requires knowing what is the value of the linear combination of no vectors at all, and that value is the zero vector. As a consequence, the zero vector always is a linear combination of whatever set$~S$ of vectors you specify. Using $n=0$ is essential when $S$ itself the empty set, in which case the zero vector is the only linear combination that one can form.

The formulation is not very clear about whether one could select the same vector more than once when forming a linear combination (it depends on whether one reads a finite "number of vectors" as a finite set or as a finite sequence). However it does not matter since allowing such repetition does not allow any more linear combinations to be formed (one can use $ax+bx=(a+b)x$ to reduce the number of occurrences of the same vector $x$ until no more repetitions occur), and in practice it is most convenient to not forbid it. For instance this allows seeing without any complications that the sum of two linear combinations from the set $S$ is again a linear combinations from the set $S$.

The definition of linear of independence uses a slightly different view on linear combinations, as is it not so much about which vectors are or are not linear combinations of elements of$~S$, but about specific linear combinations (which may be trivial combinations or not). To this end a specific linear combination of elements of$~S$ is determined by specifying for each element of$~S$ a corresponding coefficient, which association must be so that it associates a nonzero scalar only to a finite number of elements of$~S$. The value of such a linear combination is found by forming $a_1x_1+\cdots+a_nx_n$ where the sequence of vectors $x_1,\ldots,x_n$ contains each vector of$~S$ with a nonzero associated scalar exactly once, and each $a_i$ is the scalar associated to$~x_i$. A specific linear combination is trivial if the scalars associated to the vectors are all$~0$. The set$~S$ is defined to be linearly independent if the only specific linear combination of elements of$~S$ whose value is the zero vector is the trivial linear combination.

Note that this formulation arranges for the same vector to be used only once in a specific linear combination. It must do so, or else a nonzero set could never be linearly independent, because one could take any vector $x$ from the set and form for instance $1x+(-3)x+2x$ as a "nontrivial linear combination" with value the zero vector. Talking as you suggest about a "linear combinations of its distinct vectors" might convey the right idea, but does not make any precise sense (the is no definition of what such a phrase means).

You final question "if a set is linear independent, does that imply the set contains distinct vectors" is pointless, since elements of a set are always distinct: a set cannot contain a given value more than once as element. Your question would make sense if instead it was talking about sequences of families of vectors (since these may have repeated occurrences of the same values). And the answer then is that when there is at least one repetition, the sequence/family is never independent, as the difference between two instances of the same vector would be a nontrivial specific linear combination whose value is the zero vector. By the same token, and set/sequence/family containing the zero vector cannot be linearly independent (taking that vector with scalar $1$ would be a nontrivial linear combination with zero value).

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