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True/false: There are $4$ linearly independent vectors $v_1,v_2,v_3,v_4 \in \mathbb{R}^3$.

I think the statement is true.

Let's take as example (this is my own example, not of the task!!):

$$a\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}+b\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}+c\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}+d\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$$

We get that $a=b=0$

Now if $c=0$, then $d=0$, but of course we cannot know that in this case. We have to know it..?

Anyway you can see I'm not sure : /

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    $\begingroup$ What about if $c=1$? $\endgroup$ – Sha Vuklia Mar 16 '17 at 10:05
  • $\begingroup$ @ShaVuklia Then the vectors would be linearly dependent because we would have $1 = -1$, and we have that $a=b=0$ $\endgroup$ – tenepolis Mar 16 '17 at 10:08
  • $\begingroup$ @tenepolis: are you familiar with the dimension of a vector space? $\endgroup$ – Student Mar 16 '17 at 10:16
  • $\begingroup$ @student No. But I believe the dimension of the vectors are $4$ but we are in dimension $3$ and there is some kind of mismatch.. So it's impossible they are linearly independent..? $\endgroup$ – tenepolis Mar 16 '17 at 10:17
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Vectors are linearly independent if their weighted sum (linear combination) is zero if and only if all the weights are $0$s.

This is not the case in the OP:

$$0\cdot\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}+0\cdot\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}+(-1)\cdot\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}+(+1)\cdot\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}.$$

That is, the vectors above are not linearly independet.

So, the statement is false.

Above, we only showed that the example four-tuple was linearly dependent. But, it is true in general that in $R^3$ only three (or less) linearly independent vectors can be singled out.

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    $\begingroup$ it was stated in the answer informally, that there are only 3 linearly independent vectors in $\mathbb{R}^3$ but it is actually an elementary theorem of linear algebra that any set of $m$ vectors in $\mathbb{R}^n$ is linearly dependent if $m>n$. $\endgroup$ – Nox Mar 16 '17 at 10:17
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Assume $\vec v_1,\vec v_2,\vec v_3$ are linearly independent in $\mathbb{R^3}$. Then they form a basis, and we have that: $$a\vec v_1+b\vec v_2+c\vec v_3=\vec x$$

for any $\vec x\in\mathbb{R^3}$ and for some $a,b,c\in\mathbb{R}$, and therefore includes any $\vec v_4\in\mathbb{R^3}$.

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  • $\begingroup$ he might not know that any basis has the same size $\endgroup$ – enthdegree Mar 16 '17 at 17:35
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In your choice: $v_3=v_4$. Hence $v_1,v_2,v_3,v_4$ are linearly dependent:

$0=v_3-v_4$.

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  • $\begingroup$ I also like to know if the statement is true or false, please let me know if you know. $\endgroup$ – tenepolis Mar 16 '17 at 10:10
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    $\begingroup$ The statement is false ! In $ \mathbb R^3$ we have: $4$ vecors are always linearly dependent. $\endgroup$ – Fred Mar 16 '17 at 10:16

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