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

Question

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 : /

Answer 1

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.

Answer 2

In your choice: $v_3=v_4$. Hence $v_1,v_2,v_3,v_4$ are linearly dependent:

$0=v_3-v_4$.

Answer 3

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 $\vec v_4\in\mathbb{R^3}$.