Find a three independent vectors u, v, w that each lie in N(A), the null space of A. Let $A=\begin{bmatrix}0& 0& 0& 0 \\ 3& 9& 3& 9\end{bmatrix}$.
How should I figure this out? I know the first column has the Pivot and the other three columns have free variables. It has rank $1$. Will this be handy in finding the solution?
 A: let us take out  $3$ out of bracket we will get columns as 
$v_1=(0;1),  v_2=(0,3),v_3=(0;1) $ and    
$v_4=(0;3)$
independent means that if
$c_1*v_1+c_2*v_2+...+c_n*v_n=0$
then  $c_1=c_2=...=c_n=0$
now  you may figure  out which $x_1,x_2,x_3,x_4$ can be solution of this matrix
two of them can be


x =

     3
    -1
     0
     0

and
y=[1;0;-1;0]

y =

     1
     0
    -1
     0

to proof that this vectors are independent,let us solve
$c_1*(3,-1,0,0)+c_2(1,0,-1,0)=0$
we get
$3*c_1+c_2=0$
$-c_1+c_2*0=0$
$c_1*0+c_2*(-1)=0$
$c_1*0+c_2*0=0$
we can see that  $c_1=c_2=0$
thus they are independent
A: Do you see why $(3,-1,0,0)$, $(1,0,-1,0)$ and $(3,0,0,-1)$ are such vectors?
A: $x$ is in the null space when $Ax=0$, so I think it's best to start trying to solve $Ax=0.$ The equation
$$\begin{align}
\begin{bmatrix}0& 0& 0& 0 \\ 3& 9& 3& 9\end{bmatrix}\begin{bmatrix} x_1 \\ x_2\\x_3\\x_4\end{bmatrix}=&\begin{bmatrix} 0 \\ 0\end{bmatrix}\end{align}$$
implies that the only condition that $x$ needs to fulfill in order to be in the null space is that $3x_1+9x_2+3x_3+9x_4=0$ or $$x_1+3x_2+x_3+3x_4=0.$$ Now you only need to come up with three l.i. vectors which satisfy that equation. There are probably many methods to do this; I will show two:
First method
You have four parameters to specify and only one constraint, so you have freedom to be a bit arbitrary in specifying the parameters. A practical way to take advantage of this freedom is to set some of them equal to $0$ or $1$. For example if $x_1=x_2=0$ the equation becomes $x_3+3x_4=0$. There is still one free dimension, so set $x_4=1$ and this determines $x_3=-3.$ Thus, the first vector in the null space is $$v_1=(0,0,-3,1).$$ Similarly, forcing $x_3=x_4=0$ and $x_2=1$ determines a second vector in the null space $v_2=(-3,1,0,0),$ and $x_2=x_4=0,\,x_1=1$ determines a third one $v_3=(1,0,-1,0).$ You have to make sure they are l.i. afterwards.
Second method
And my favorite, is to solve for one of the variables: $$x_1=-3x_2-x_3-3x_4,$$ so that a solution will be of the "parametrized" form $$\begin{bmatrix}-3x_2-x_3-3x_4 \\ x_2 \\ x_3 \\ x_4\end{bmatrix}.$$ Now just do the "set equal to $0/1$ thing". Choose 


*

*$x_2=1, \; x_3=x_4=0$ for $w_1=(-3,1,0,0)$,

*$x_4=1, \; x_2=x_3=0$ for $w_2=(-1,0,1,0)$, and

*$x_3=1, \; x_2=x_4=0$ for $w_3=(-3,0,0,1)$.


These will obviously be l.i., because any linear combination of them with nonzero coefficients will have nonzero second, third or fourth coordinates.
