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I needed clarification on a linear algebra question that I had:

Given the matrices $v_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}, $ $v_2 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}$ and $v_3 = \begin{bmatrix} 1 \\ 1 \\ -1 \\ \end{bmatrix}$,

1) How many vectors does the set {${v_1, v_2, v_3}$} have?

2) How many vectors are in Span{$v_1, v_2, v_3$}?

I think the answer to #1 is 3, simply because there are three matrices, and the answer to #2 is infinite, since there are an infinite number of linear combinations that can be made using these vectors.

I am uncertain of these answers, though.

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    What does $<v_1,v_2,v_3>$ mean? What is the underlying field? –  Mar 06 '17 at 22:56
  • If they are linearly independent, your answers are true so checking that will lead the result. – Ninja Mar 06 '17 at 22:57
  • yes they are linearly independent since they are not multiples of each other –  Mar 06 '17 at 23:00
  • < .. > is notation for set here. i edited it –  Mar 06 '17 at 23:00
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    @Ninja: this is even the case whenever they are not linearly independent: none of the 3 is the zerovector, so even if they are linearly independent, it would be a space of dimension at least 1, hence the number of vectors depends on the considered field. (If it is $\mathbb{R}$, then $\text{span}{v_1}$ still contains infinitely many elements, if it is a finite field, then it is the number of elements in the field). For the first question: even if they are linearly dependent, they are still different, hence the answer to question 1 is always 3. – Student Mar 06 '17 at 23:06
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    @stackofhay42 For 3+ vectors, linearly dependent does NOT mean "multiple of". Consider $(1, 0), (1, 1), (0, 1)$ for an example of three that are dependent but none is a multiple of the others. – btilly Mar 06 '17 at 23:07
  • @Student: Unless $-1=1$, i.e. in characteristic $2$. – Berci Mar 06 '17 at 23:09
  • So what I have is correct ? –  Mar 06 '17 at 23:12
  • @Berci: nicely done, didn't think of that field :) anyway, my point is that in the case where we have a field of characteristic $0$, linear dependency does not chance the number of elements in the span. (I think I may assume that, given the content of this question, the vectorspace is considered over $\mathbb{R}$ or $\mathbb{C}$). at Stackofhay42: Unless you are working in some field with non zero characteristic, your answer is correct! Perhaps you should mention over what field you consider your vectorspace. – Student Mar 06 '17 at 23:14
  • this would be over ℝ I believe ? –  Mar 06 '17 at 23:15
  • @stackofhay42: It is probably mentioned somewhere... Does the exercise state its a real vectorspace? (I do suspect that you are not working over finite fields, so it does not really matter if it is real or complex, except for previously mentioned special cases) – Student Mar 06 '17 at 23:18

1 Answers1

-1

Construct the matrix $$ \mathbf{A} = \left[ \begin{array}{ccc} v_{1} & v_{2} & v_{3} \end{array}\right] = \left[ \begin{array}{rrr} \phantom{-}1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right] . $$ Because $\det \mathbf{A} = 4 \ne 0$, the vector set is linearly independent.

The span of the vector set is $$ \text{span} \left\{ v_{1}, v_{2}, v_{3} \right\} = \mathbb{R}^{3}. $$ These vectors are a linearly independent span, also called a minimal spanning set for $\mathbb{R}^{3}$.

dantopa
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