So I have these $4$ vectors (they are column vectors, but I can't put them in the right form , somebody please help)
$v_1= (1,1,0,0)$
$v_2=(1,-1,0,0)$
$v_3=(0,2,0,0)$
$v_4=(0,2,1,0)$
The professor said that only $v_1,v_2,$ and $v_4$ are linearly independent.
He remarks by saying that that $v_3 = v_1 − v_2$; since $v_1$ and $v_2$ are clearly independent and both have $0$ as their third coordinate while $v_4$ has a non-zero third coordinate, a set of three linearly independent vectors is clearly provided by $\{v_1, v_2, v_4\}$.
I did not understand this process. Can somebody please help me understand why only $3$ vectors out of $4$ are linearly independent? Thank you!
I know that vectors are L.I. if they have only the trivial solution , but I'm not sure how to apply this here.
Please explain as simply as you can too, if that is possible.
(*Since you are a new user I shall try to explain. But is best if you show in this forum what you have done so that we can guide you.)
Linear Independence: $v_1,v_2,\cdots,v_n$ are linearly independent if the only solution to the equation $a_1v_1+a_2v_2+\cdots+a_nv_n=0$ is $a_1=a_2=\cdots=a_n=0$.
So here we have $a_1v_1+a_2v_2+\cdots+a_nv_n=0 \implies a_1\begin{bmatrix}1\\1\\0\\0\end{bmatrix}+a_2\begin{bmatrix}1\\-1\\0\\0\end{bmatrix}+a_3\begin{bmatrix}0\\2\\0\\0\end{bmatrix}+a_4\begin{bmatrix}0\\2\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$ which gives us four equations $$a_1+a_2=0\\a_1-a_2+2a_3+2a_4=0\\a_3=0.$$ Using $a_3=0,a_1=-a_2$ in the second equation gives you $a_3=a_4$. So we have got a non-zero solution $(-a_2,a_2,0,a_2)$ for $a_1v_1+a_2v_2+\cdots+a_4v_4=0 $. Thus the four vectors $\{a_1,a_2,a_3,a_4\}$ are liearly dependent.
You may try to show linear independence or dependence with any set of $3$ vectors from the above $4$ vectors to understand further.