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

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 : /

• What about if $c=1$? – Sha Vuklia Mar 16 '17 at 10:05
• @ShaVuklia Then the vectors would be linearly dependent because we would have $1 = -1$, and we have that $a=b=0$ – tenepolis Mar 16 '17 at 10:08
• @tenepolis: are you familiar with the dimension of a vector space? – Student Mar 16 '17 at 10:16
• @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..? – tenepolis Mar 16 '17 at 10:17

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.

• 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 $m>n$. – Nox Mar 16 '17 at 10:17

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

• he might not know that any basis has the same size – Christian Chapman Mar 16 '17 at 17:35

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

$0=v_3-v_4$.

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