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Prove the following statement.

Let $u$, $v$, $w$ ∈ $R^{2}$ be any three two-dimensional vectors. Then $u$, $v$, $w$ are linearly dependent.

What i tried

Proving by contradiction

I let the vectors $u$, $v$, $w$ be linearly independent.

Then the vectors can be written as a linear combination

$$c_{1}u+c_{2}v+c_{3}w=0$$

Since we know they are linearly independent and they span $R^{3}$. That means they are a basis of $R^{3}$. But our assumption mentions $u$, $v$, $w$ ∈ $R^{2}$ which is a contradiction. Hence proving the original statement.

Is my proof correct. Could anyone explain. Thanks

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    $\begingroup$ The set $\mathbb{R}^2$ is composed of objects that are fundamentally distinct from objects in $\mathbb{R}^3$. Therefore, if $u \in \mathbb{R}^2$, then $u=(x,y)$ for some $x,y \in \mathbb{R}$. You cannot then claim that $u \in \mathbb{R}^3$, because $u$ is not an ordered triple! $\endgroup$ – David Kraemer Oct 5 '15 at 17:56
  • $\begingroup$ Tell us what you were able to deduce so we can check if you understood. Cheers! $\endgroup$ – Kevin Zakka Oct 5 '15 at 18:13
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Use the following:

  • Form the non-square matrix A $(m \times n)$ such that the column vectors of A are the vectors of S (i.e $A_i = v_i$ i = 1, 2, ... , n)

  • What can you say about the number of free variables of A?

Keep in mind that a linearly dependent set has at least 1 non trivial solution, that is, infinitely many solutions.


On a related note, there is a theorem that states that if $S = \{v_1, v_2, ... ,v_n\} $ is a set of nonzero vectors in $R^m$ then S is linearly dependent.

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  • $\begingroup$ I let $A$ be a $2$ by $3$ matrix and i assume $A$ to be linearly independent. Hence the solution of $Ax=0$ must only have the trivial solution. Hence only a unique solution. If we row reduce $A$ to row echlon form we notice that $A$ have at most $2$ leading entries. Hence there will be a free column and we kno that the solution $Ax=0$ will have infinetly many solutions. Hence a contradiction. Is this correct Thanks? $\endgroup$ – ys wong Oct 5 '15 at 18:22
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    $\begingroup$ Great! I'd like to point out that you do not need to assume A is l.i. and then try to prove a contradiction. Directly write $c_1 v_1 + ... + c_nv_n$ into its matrix form which is the homogeneous system $Ac = 0$. Then conclude directly that since 3 > 2, there is at least one free variable thus the system does not have a unique solution thus your initial three vectors are l.d. $\endgroup$ – Kevin Zakka Oct 5 '15 at 18:37
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It would be far easier to use the fact that the dimension of the vector space is the largest possible number of linearly independent vectors, and since $3>2$ they must be linearly dependent.

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    $\begingroup$ This is circular: This argument presupposes that one knows that $\dim_{\Bbb R} (\Bbb R^2) = 2,$ but by definition (of dimension) the statement to proved is that $\dim_{\Bbb R} (\Bbb R^2) \leq 2$. $\endgroup$ – Travis Oct 5 '15 at 18:00
  • $\begingroup$ @Travis This is not circular. One can easily furnish a basis for $\mathbb{R}^2$, and then use the fact that if one set $A$ is linearly independent and one set $B$ generates the vector space, then $\#A \leq \#B$, which can be proved independently (see my answer here math.stackexchange.com/questions/1249040/… ) $\endgroup$ – Aloizio Macedo Oct 5 '15 at 19:26

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